Solving Trigonometric Equations - Trigonometry

Card 0 of 264

Solve for $\theta$ : $2 \sin ^2 (3 \theta ) = 1$

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There are multiple solution paths. We could subtract 1 from both sides and use the quadratic formula with and . Or we could solve using inverse opperations:

$2 \sin ^2 (3 \theta) = 1$ divide both sides by 2

$\sin ^2 (3 \theta ) = $\frac{1}{2}$$ take the square root of both sides

$\sin (3 \theta ) = \pm $\frac{1}$ { \sqrt 2 }$

The unit circle tells us that potential solutions for $3 \theta$ are $\frac{\pi }{4}$, $\frac{3 \pi }{4}$ , $\frac{5 \pi }{4}$ , $\frac{7 \pi }{4}$ .

To get our final solution set, divide each by 3, giving:

$\frac{ \pi }{12}$, $\frac{\pi }{4 }$, $\frac{5 \pi }{12}$, $\frac{ 7 \pi }{12 }$ .

Solve for $\theta$ : $4 \sin ^2( \theta + $\frac{ \pi }{3 }$) + 2 = 5$

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This problem has multiple solution paths, including subtracting 5 from both sides and using the quadratic formula with . We can also solve using inverse opperations:

$4 \sin ^2 (\theta + $\frac{ \pi }{3}$ ) + 2 = 5$ subtract 2 from both sides

$4 \sin ^2 (\theta + $\frac{ \pi }{3}$ ) = 3$ divide both sides by 4

$\sin ^ 2 (\theta + $\frac{ \pi }{3}$ ) = $\frac{3}{4}$$ take the square root of both sides

$\sin (\theta + $\frac{ \pi }{3}$ ) = \pm $\frac{\sqrt3}{2}$$

If the sine of an angle is $\pm $\frac{\sqrt 3 }{2}$$ , that angle must be one of $\frac{ \pi }{3}$ , $\frac{2 \pi }{3}$ , $\frac{4 \pi }{3}$ , $\frac{5 \pi }{3}$ . Since the angle is $\theta + $\frac{ \pi }{3}$$ , we can get theta by subtracting $\frac{ \pi }{3}$ :

$$\frac{\pi }{3}$ - $\frac{ \pi }{3}$ = 0$

$\frac { 2 \pi }{3} - $\frac{ \pi }{3}$ = $\frac{ \pi }{3}$$

$$\frac{ 4 \pi }{3}$ - $\frac{ \pi }{3}$ = $\frac{ 3 \pi }{3}$ = \pi$

$\frac{ 5 \pi }{3}$ - $\frac{ \pi }{3}$ = $\frac{ 4 \pi }{3}$

Solve the following equation for $0\leq x<2\pi$ .

$\sin\left($\frac{x}{2}\right)=\cos x -1$

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$\sin\left($\frac{x}{2}\right)=\cos x -1$ ; We start by substituting a new variable. Let $u=\frac{x}{2}$$ .

$\sin u=\cos 2u-1$ ; Use the double angle identity for cosine

$\sin u = 1-2\sin ^2 u-1$ ; the 1's cancel, so add $2\sin ^2u$ to both sides

$2\sin ^2u+\sin u=0$ ; factor out a $\sin u$ from both terms.

$\sin u(2\sin u+1)=0$ ; set each expression equal to 0.

$\sin u =0$ or $2\sin u +1=0$ ; solve the second equation for sin u.

$\sin u = 0$ or $\sin u = -$\frac{1}{2}$$ ; take the inverse sine to solve for u (use a unit circle diagram or a calculator)

$\frac{x}{2}$=u=0, $\frac{7\pi}{6}$, $\frac{11\pi}{6}$ ; multiply everything by 2 to solve for x.

$x=0, $\frac{7\pi}{3}$, $\frac{11\pi}{3}$$ ; Notice that the last two solutions are not within our range $0\leq x<2\pi$ . So the only solution is .

Solve the equation below for greater than or equal to and strictly less than .

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Recall the values of for which . If it helps, think of sine as the values on the unit circle. Thus, the acceptable values of would be 0, 180, 360, 540 etc.. However, in our scenario .

Thus we have and .

Any other answer would give us values greater than 90. When we divide by 4, we get our answers,

and .

$f(x)=2$\sin{(2x)}$-2\sin(x)\cos(x)$

Find the three smallest positive roots of the above equation.

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By the double angle identity, we can find

$-2\sin(x)\cos(x)=-\sin(2x)$

$f(x)=2\sin(2x)-\sin(2x)=\sin(2x)$

So to get the zeros, solve:

$\sin(2x)=0$

This means that any number that when doubled equals a multiple of 180 degrees is a zero. In this case that includes

$0, \pm $90^{\circ}$, $\pm180^{\circ}$, $\pm270^{\circ}$, etc.$

But the question asks for the smallest positive roots which excludes the negative and zero roots, leaving 90, 180, 270

Which of the following is NOT a solution to the equation below such that $0\leq x<2\pi$ ?

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Given the multiple choice nature of the problem, the easiest way to solve would be to simply plug in each answer and find the one that does not work.

However, we want to learn the math within the problem. We begin solving the equation by factoring

We then divide.

We then must remember that our left side is equivalent to something simpler.

We can therefore substitute.

$cos(2x)=\frac{1}{2}$$

We then must consider the angles whose cosine is $\frac{1}{2}$ . The two angles within the first revolution of the unit circle are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ , but since our angle is , we need to consider the second revolution, which also gives us $\frac{7\pi}{3}$ and $\frac{11\pi}{3}$ .

But since is equal to each of these angles, we must divide them by 2 to find our answers. Therefore, we have

$x=\frac{\pi}{6}$,$\frac{\5\pi}{6}$,$\frac{7\pi}{6}$,$\frac{11\pi}{6}$$

Therefore, there is only one answer choice that does not belong.

Which of the following is a solution to the following equation such that $0\leq x<2\pi?$

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We begin by getting the right side of the equation to equal zero.

Next we factor.

We then set each factor equal to zero and solve.

or

$sin(x)=-$\frac{1}{2}$$

We then determine the angles that satisfy each solution within one revolution.

The angles $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ satisfy the first, and $\frac{\pi}{2}$ satisfies the second. Only $\frac{11\pi}{6}$ is among our answer choices.

Solve the following equation. Find all solutions such that $0\leq x <2\pi$ .

$2\sin x = 1$

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$2\sin x=1$ ; Divide both sides by 2 to get

$\sin x=\frac{1}{2}$$ ; take the inverse sine on both sides

; the left side reduces to x, so

$x=\sin ^{-1}\bigg($\frac{1}{2}$\bigg)$

At this point, either use a unit circle diagram or a calculator to find the value.

Keep in mind that the problem asks for all solutions between and $2\pi$ .

If you use a calculator, you will only get $\frac{\pi}{6}$ as an answer.

So we need to find another angle that satisfies the equation $\sin x=\frac{1}{2}$$ .

Solve the following equation. Find all solutions such that $0\leq x<2\pi$ .

$\sin 2x=4\cos x$

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$\sin 2x=4\cos x$ ; First use the double angle identity for $\sin 2x$ .

$2\sin x \cos x = 4\cos x$ ; divide both sides by 2

$\sin x \cos x=2\cos x$ ; subtract the $2\cos x$ from both sides

$\sin x \cos x-2\cos x=0$ ; factor out the $\cos x$

$\cos x(\sin x-2)=0$ ; Now we have the product of two expressions is 0. This can only happen if one (or both) expressions are equal to 0. So let each expression equal 0.

$\cos x= 0$ or $\sin x-2=0$ ;

$\cos x = 0$ or $\sin x=2$ ; Take the inverse of each function for each expression.

$x=\cos ^{-1}(0)$ or $x=\sin ^{-1}(2)$ ; The second equation is not possible so gives no solution, but the first equation gives us:

$x=\frac{\pi}{2}$; $\frac{3\pi}{2}$$

Solve the following equation for $0\leq x<2\pi$ .

$\sin 2x=\cos 2x$

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The fastest way to solve this problem is to substitute a new variable. Let .

The equation now becomes:

$\sin u=\cos u$

So at what angles are the sine and cosine functions equal. This occurs at

$u=\frac{\pi}{4}$; $\frac{5\pi}{4}$; $\frac{9\pi}{4}$; $\frac{13\pi}{4}$$

You may be wondering, "Why did you include

$\frac{9\pi}{4}$; $\frac{13\pi}{4}$ if they're not between and $2\pi$ ?"

The reason is because once we substitute back the original variable, we will have to divide by 2. This dividing by 2 will bring the last two answers within our range.

$2x=\frac{\pi}{4}$; 2x=\frac{5\pi}{4}$; 2x=\frac{9\pi}{4}$; 2x=\frac{13\pi}{4}$$

Dividing each answer by 2 gives us

$x=\frac{\pi}{8}$; $\frac{5\pi}{8}$; $\frac{9\pi}{8}$; $\frac{13\pi}{8}$$

Solve the equation for $0\leq x<2\pi$ .

$\cos 4x = \cos 2x$

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We begin by substituting a new variable .

$\cos 2u=\cos u$ ; Use the double angle identity for $\cos 2u$ .

$2\cos ^2 u-1=\cos u$ ; subtract the $\cos u$ from both sides.

$2\cos ^2 u-\cos u-1=0$ ; This expression can be factored.

$(2\cos u+1)(\cos u-1)=0$ ; set each expression equal to 0.

$2\cos u+1=0$ or $\cos u -1=0$ ; solve each equation for $\cos u$

$\cos u = -$\frac{1}{2}$$ or $\cos u = 1$ ; Since we sustituted a new variable we can see that if $0\leq x<2\pi$ , then we must have $0\leq 2x < 4\pi$ . Since , that means $0\leq u<4\pi$ .

This is important information because it tells us that when we solve both equations for u, our answers can go all the way up to $4\pi$ not just $2\pi$ .

So we get

$2x = u=0; $\frac{2\pi}{3}$; $\frac{4\pi}{3}$; 2\pi; $\frac{8\pi}{3}$; $\frac{10\pi}{3}$;$ Divide everything by 2 to get our final solutions

$x=0; $\frac{\pi}{3}$; $\frac{2\pi}{3}$; \pi; $\frac{4\pi}{3}$; $\frac{5\pi}{3}$;$

Solve the following equation for $0\leq x<2\pi$ .

$4\cos ^2x=3$

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$4\cos ^2x=3$ ; First divide both sides of the equation by 4

$\cos ^2x=\frac{3}{4}$$ ; Next take the square root on both sides. Be careful. Remember that when YOU take a square root to solve an equation, the answer could be positive or negative. (If the square root was already a part of the equation, it usually only requires the positive square root. For example, the solutions to are 2 and -2, but if we plug in 4 into the function $f(x)=$\sqrt{x}$$ the answer is only 2.) So,

$\cos x = \pm $\frac{\sqrt{3}$}{2}$ ; we can separate this into two equations

$\cos x = $\frac{\sqrt{3}$}{2}$ and $\cos x =- $\frac{\sqrt{3}$}{2}$ ; we get

$x=\frac{\pi}{6}$; $\frac{11\pi}{6}$$ and $x=\frac{5\pi}{6}$; $\frac{7\pi}{6}$$

Solve the equation for $0\leq x<2\pi$ .

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; Divide both sides by 3

; Take the square root on both sides. Just as the previous question, when you take a square root the answer could be positive or negative.

$\tan x=\pm $\frac{1}{\sqrt{3}$}$ ; This can be written as two separate equations

$\tan x=\frac{1}{\sqrt{3}$}$ and $\tan x=-$\frac{1}{\sqrt{3}$}$ ; Take the inverse tangent

$x=\frac{\pi}{6}$; $\frac{7\pi}{6}$$ and $x=\frac{5\pi}{6}$; $\frac{11\pi}{6}$$

Solve the following equation for $0\leq x<2\pi$ .

$\tan ^2x+2\tan x+1=0$

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$\tan ^2x+2\tan x+1=0$ ; The expression is similar to a quadratic expression and can be factored.

$(\tan x+1)(\tan x+1)=0$ ; set both expressions equal to 0. Since they are the same, the solutions will repeat, so I will only write it once.

$\tan x+1=0$

$\tan x=-1$ ; take the inverse tangent on both sides

$x=\tan ^{-1}(-1)$

$x=\frac{3\pi}{4}$; $\frac{7\pi}{4}$$

Solve the following equation for $0\leq x<2\pi$ .

$2\sin ^2x=3\cos(2x)$

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$2\sin ^2x=3\cos(2x)$ ; use the double angle identity for cosine

$2\sin ^2x=3(1-2\sin ^2x)$ ; distribute the 3 on the right side

$2\sin ^2x=3-6\sin ^2x$ ; add the $6\sin ^2x$ to both sides

$8\sin ^2x=3$ ; divide both sides by 8

$\sin ^2x=\frac{3}{8}$$ ; take the square root on both sides (Remember: it could be positive or negative)

$\sin x=\pm $\frac{3}{8}$$ ; separate into two equations and take the inverse sine

$x=\sin ^{-1}\left($\sqrt{\frac{3}{8}$} \right)$ and $x=\sin ^{-1}\left(-$\sqrt{\frac{3}{8}$} \right )$ ; Use a calculator

and (calculator will give -0.659, but that is not in our range, so add $2\pi$ to get 5.624)

The last two solutions are found using a unit circle. Since our x can be negative or positive this means that there is a corresponding value in all of the quadrants.

Similarly, we can get our last answer x= 3.801 as this is the x value in the third quadrant and is found by adding $0.659+\pi=3.801$ .

The equation $y=-6\cos 3t$ represents the motion of a weight hanging on a spring after it has been pulled 6 inches below its equilibrium point and released. The variable represents the weight's position above or below the equilibrium point. For example, means the weight is 5 inches below its equilibrium point. means the weight is 5 inches above its equilibrium point. The variable represents the time, in seconds, after the weight was released.

Find the first 2 times the weight is 3 inches below its equilibrium point.

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You are asked to find the first TWO times the the weight is 3 inches below its equilibrium point. Since it is below the equilibrium, that 3 will be negative. The y variable is used to represent position so we need to have .

$y=-3=-6\cos 3t$ ; now we solve. First, divide by -6.

$$\frac{1}{2}$=\cos 3t$ ; At this point, it may help to substitute a new variable. Once you get more practice with this type of problem, you can skip the substitution.

Let , then

$\cos u = $\frac{1}{2}$$ ; using a calculator or a unit circle diagram, find the first TWO angles that give us a cosing value of 1/2.

$u=\frac{\pi}{3}$, $\frac{5\pi}{3}$$ ; bring back the variable t

$3t=\frac{\pi}{3}$, $\frac{5\pi}{3}$$ ; divide each solution by 3

$t=\frac{\pi}{9}$, $\frac{5\pi}{9}$$ .

The equation $y=3\sin\left(4t-$\frac{\pi}{2}\right)$ represents the motion of a weight hanging on a spring after it has been pulled 6 inches below its equilibrium point and released. The variable represents the weight's position above or below the equilibrium point. For example, means the weight is 5 inches below its equilibrium point. means the weight is 5 inches above its equilibrium point. The variable represents the time, in seconds, after the weight was released.

Find the first three times the weight is 1.5 inches above its equilibrium point.

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Similar to the previous question, we set up the equation with

$1.5=3\sin\left(4t-$\frac{\pi}{2}\right)$ ; divide both sides by 3

$$\frac{1}{2}$=\sin\left(4t-$\frac{\pi}{2}\right)$ ; Use a substitution to make things easier $u=4t-$\frac{\pi}{2}$$

$$\frac{1}{2}$=\sin u$ ; take the inverse sine. Use a calculator or unit circle diagram to find the first THREE solutions (that is what the problem asked for).

$u=\frac{\pi}{6}$; $\frac{5\pi}{6}$; $\frac{13\pi}{6}$$ ; bring back the expression for u and separate your answers.

$4t-$\frac{\pi}{2}$=\frac{\pi}{6}$; 4t-$\frac{\pi}{2}$=\frac{5\pi}{6}$; 4t-$\frac{\pi}{2}$=\frac{13\pi}{6}$$ ; solve each equation.

Notice that you will have to add $\frac{\pi}{2}$ for each equation, but all the other fractions have a denominator of 6. $\frac{\pi}{2}$=\frac{3\pi}{6}$

$4t-$\frac{3\pi}{6}$=\frac{\pi}{6}$; 4t-$\frac{3\pi}{6}$=\frac{5\pi}{6}$; 4t-$\frac{3\pi}{6}$=\frac{13\pi}{6}$$ ; add $\frac{3\pi}{6}$ to both sides of each equation.

$4t=\frac{4\pi}{6}$; 4t=\frac{8\pi}{6}$; 4t=\frac{16\pi}{6}$$ ; Divide by 4 in each equation

$t=\frac{\pi}{6}$; t=\frac{2\pi}{6}$; t=\frac{4\pi}{6}$$ ; And finally simplify

$t=\frac{\pi}{6}$; t=\frac{\pi}{3}$; t=\frac{2\pi}{3}$$

Solve for . Give your answer in radians.

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We can factor the original expression as . This gives us 2 factors that we need to individually set equal to 0.

The simpler one is:

. The two values for x that give a cosine of 0 are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

Now set the second factor equal to 0:

add 1 to both sides

divide by 2

$cosx=\frac{1}{2}$$ . The two values for x that give a cosine of $\frac{1}{2}$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ .

In total we have 4 answers, $x=\frac{\pi}{2}$; $\frac{3\pi}{2}$; $\frac{\pi}{3}$; $\frac{5\pi}{3}$$

Solve for . Give your answer in radians.

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We can think of this equation as:

$sinx \cdot sinx - $\frac{sinx}{cosx}$=0$

Then we can factor it as:

$sinx[sinx-$\frac{1}{cosx}$]=0$ . Now we set both factors equal to 0 and solve.

The simpler one is . Sine is 0 when x is 0 or $\pi$ . $2\pi$ would also work, although it is just another way of saying 0.

Now we will set the second factor equal to zero:

$sinx-$\frac{1}{cosx}$=0$ add $\frac{1}{cosx}$ to both sides

$sinx = $\frac{1}{cosx}$$ multiply both sides by

$sinx\cdot cosx=1$

This must have no solution, because sine and cosine are never any larger than 1, and they are never 1 simultaneously. This means that our only answers are 0 and $\pi$ .

Solve for , giving an answer in radians.

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This expression can be factored as:

$sinx[2sinx-\sqrt3]=0$ . We can solve by setting each factor equal to 0.

gives the answers $x=0, \pi$ since sine is 0 for those values of x.

To continue solving, set the second factor equal to 0:

$2sinx-\sqrt3 = 0$ add $\sqrt3$ to both sides.

$2sinx=\sqrt3$ divide both sides by 2.

$sinx=\frac{\sqrt3}$2$

The values of x that give a sine of $\frac{\sqrt3}{2}$ are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ , so this equation has 4 solutions.