Quadratic Formula with Trigonometry - Trigonometry

Card 0 of 112

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$

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

What are the zeros of the function listed above for the interval $[0,2\pi]$ .

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When the quadratic formula is applied to the function, it yields

$$\frac{-b\pm $\sqrt{b^2$-4ac}$}{2a}=\frac{-5\pm \sqrt{25-16}$}{4}=\frac{5\pm3}{4}$= =-$\frac{1}{2}$, -2$

So those are the zeros for sine, but sine has a minimum of -1, so -2 is out. For -1/2, sine achieves that twice in a cycle, at π+π/6 and 2π-π/6. So while -π/6 is true, it is not correct since it is not in the given interval.

Therefore on the given interval the zeros are:

$\frac{7\pi}{6}$, $\frac{11\pi}{6}$

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

$3\sin x=\cos 2x$

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$3\sin x=\cos 2x$ ; Use the double angle identity for cosine.

$3\sin x=1-2\sin ^2x$ ; Move everything to the left side of the equation.

$2\sin ^2x+3\sin x-1=0$ ; This is a quadratic-like expression that cannot be factored. We must use the quadratic formula. It may be helpful to see this if you replace $\sin x$ with , so it becomes:

Recall the quadratic formula $y=\frac{-b\pm $\sqrt{b^2$-4ac}$}{2a}$

plug in .

We now have

$\sin x = $\frac{-3\pm \sqrt{17}$}{4}$ ; Separate this into two equations and take the inverse sine.

$x = \sin ^{-1}\left( $\frac{-3+\sqrt{17}$}{4}\right)$ or $x = \sin ^{-1}\left( $\frac{-3-\sqrt{17}$}{4}\right )$

The first equation gives us . Using the unit circle as we did in previous problems, we can find a second answer from this which is . The second equation will not give us a solution.

Solve the following trigonometric equation:

for $x\in [0,$\frac{\pi}{2}$]$

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Since can be written as:

. We can't have .

Therefore . This means that $x=2k\pi,$ where k is an integer.

since $\in[0,\p$\frac{\pi}{2}$]$ . We have x=0 is the only number that satisfies this property.

Solve each equation over the domain (answer in degrees).

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Rearrange the problem,

Over the interval 0 to 360 degrees, cosx = 1/2 at 60 degrees and 300 degrees.

Solve the equation over the interval $0 < x < 2\pi$

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First, get the equation in terms of one trig function. We can do this by substituting in using the Pythagorean Identity for .

Then we have .

Bring all the terms to one side to find that .

We can factor this quadratic to .

This means that .

The only angle value for which this is true is $\frac{3\pi}{2}$ .

Solve for :

.

Give your answer as a positive angle measure.

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Use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=4, b=1, and c=-1:

$sinx = $$\frac{-1\pm\sqrt{(-1)^2$-(4\cdot 4\cdot -1)}$}{2\cdot 4}$ simplify

$sinx = $\frac{-1\pm\sqrt{1+16}$}{8}$

$sinx = $\frac{-1\pm\sqrt{17}$}{8}$

the square root of 17 is about 4.123. This gives two potential answers:

. We can solve for x by evaluating both and . The first gives an answer of . Add this to 360 to get that as a positive angle measure, . If this has a sine of -0.64, so does its reflection over the y-axis, which is .

The second gives an answer of . If that has a sine of 0.39, then so does its reflection over the y-axis, which is .

Solve for , giving your answer as a positive angle measure:

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First, re-write the equation so that it is equal to zero:

Now we can use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=1, b=2, and c=-3:

$sinx = $$\frac{-2\pm\sqrt{(-2)^2$-(4\cdot 1\cdot -3)}$}{2\cdot 1}$ simplify

$sinx = $\frac{-2\pm\sqrt{4+12}$}{2}$

$sinx = $\frac{-2\pm\sqrt{16}$}{2}$

$sinx=\frac{-2\pm4}{2}$$

This gives two potential answers:

$sinx=\frac{-2+4}{2}$=\frac{2}{2}$=1$

and

$sinx=\frac{-2-4}{2}$=\frac{-6}{2}$=-3$

Sine must be between -1 and 1, so there are no values of x that would give a sine of -3. The only solution that works is . The only angle measure that has a sine of 1 is .

Solve for $\theta$ : $4 \cos ^2 \theta - 4 \cos \theta - 3 = 0$

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To solve, use the quadratic formula with and $\cos \theta$ where x would normally be:

$\cos \theta = $\frac{4 \pm \sqrt{16 -4(4)(-3)}$}{2(4)} = $\frac{4 \pm 8}{8 }$$

This gives us two potential answers:

$\cos \theta = $\frac{4 + 8 }{8 }$ = $\frac{12}{8 }$ = 1.5$ since this number is greater than 1, it is outside of the domain for cosine and won't give us any solutions.

$\cos \theta = $\frac{4 - 8 }{8}$ = $\frac{ -4}$ { 8 } = $\frac{-1}{2}$$

Consulting the unit circle, the cosine is $-$\frac{1}{2}$$ when $\theta = $\frac{2 \pi }{3}$ , $\frac{ 4 \pi }{3}$$

Solve for $\theta$ : $2 \cos ^ 4 \theta - 9 \cos ^2 \theta + 4 = 0$

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To start solving, first realize that this is a quadratic with "x" as $\cos ^2 \theta$ :

$2[\cos ^2 \theta $]^2$ - 9 $[\cos^2$ \theta] + 4 = 0$

We can solve using the quadratic formula:

One potential solution: $\cos ^2 \theta = $\frac{9 + 7 }{4 }$ = $\frac{16 }{4}$ = 4$

Taking the square root gives: $\cos \theta = 2$ , but 2 is outside the range of cosine, so that won't work.

The other potential solution: $\cos ^2 \theta = $\frac{ 9 - 7 }{4}$ = $\frac{ 2}{4}$ = $\frac{1}{2}$$

Taking the square root gives: $\cos \theta = \pm $\frac{1 }{\sqrt 2 }$$

Consulting the unit circle, $\theta = $\frac{ \pi }{4}$ , $\frac{3 \pi }{4}$ , $\frac{5 \pi }{4}$ , $\frac{ 7 \pi }{4}$$

Solve for $\theta$ : $5 \sin ^ 2 (2 \theta ) - 13 \sin (2 \theta ) - 6 = 0$

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To solve, first use the quadratic formula:

$\sin ( 2\theta )= $\frac{ 13 \pm $\sqrt{13^2$- 4(5)(-6)}$}{2(5)} = $\frac{ 13 \pm 17}{10}$$

This gives us two potential solutions:

$\sin ( 2 \theta ) = $\frac{ 13 + 17 }{ 10 }$ = $\frac{ 30 }{10 }$ = 3$ that is outside the range of sine, so it won't work

$\sin ( 2 \theta ) = $\frac{ 13 - 17 }{ 10 }$ = $\frac{ -4 }{10 }$ = -0.4$

We can continue solving by taking the inverse sine:

$2 \theta = \sin ^ {-1 } (-0.4)$

using a calculator gives us

$2 \theta \approx -23.58$ , which we can convert to a positive angle measure by adding to 360:

This is just one answer for $2 \theta$ . The other angle that would work would be below , or

Since those are values for $2 \theta$ , to get our final answers divide by 2:

Solve for $\theta$ : $4 \sin ^ 2 \theta - \sin \theta - 3 = 0$

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First solve using the quadratic formula:

$\sin \theta = \frac { 1 \pm $\sqrt{1 - 4 (4)(-3)}$}{2(4)} = $\frac{ 1 \pm 7 }{ 8 }$$

This gives two potential solutions:

$\sin \theta = $\frac{ 1 + 7 }{8 }$ = $\frac{ 8}{8 }$ = 1$

The only value for $\theta$ where sine is 1 is .

$\sin \theta = $\frac{ 1 - 7 }{8 }$ = $\frac{ -6 }{ 8 }$ = -0.75$

Using a calculator, we get $\theta = \sin ^ {-1} (-0.75 ) \approx -48.59$

Adding that to 360 givesus the angle's positive value,

That's just one instance where the sine is -0.75. We also need to find the other angle below the x-axis by adding .

So our three values for theta are

Solve for $\theta$ : $2 \cos ^2 (4 \theta ) + 9 \cos (4 \theta) = 5$

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First, solve for $\cos (4 \theta)$ using the quadratic formula:

$\cos ( 4 \theta) = $\frac{ -9 \pm \sqrt{81 - 4(2)(-5)}$}{4} = $\frac{ 9 \pm 11}{4}$$

This gives two solutions:

$\cos (4 \theta ) = $\frac{ -9 -11 }{ 4}$ = $\frac{ -20 }{4}$ = -5$ this is outside of the range of cosine so it will not work.

$\cos (4 \theta ) = $\frac{ -9 + 11}{ 4 }$ = $\frac{ 2 }{4 }$ = $\frac{ 1}{2}$$

Consulting the unit circle tells us that $4 \theta = $60^o$$ or . To get our final answers, just divide these by 4:

$\theta = $15^o$, $75^o$$

Solve for $\theta$ : $10 \cos ^2 ( \theta + 1 ) + 19 \cos (\theta + 1 ) - 15 = 0$

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First, solve for $\cos ( \theta + 1 )$ by using the quadratic formula:

$\cos (\theta + 1 ) = $\frac{ - 19 \pm \sqrt{ 361 - 4 (10)(-15)}$}{20 } = $\frac{ -19 \pm 31 }{20}$$

This gives two solutions:

$\cos ( \theta + 1) = $\frac{ - 19 - 31 }{ 20 }$ = $\frac{ -50 }{20 }$ = -2.5$ this is outside of the range for cosine, so that does not work as a solution

$\cos ( \theta + 1 ) = $\frac{ -19 + 31 }{ 20 }$ = $\frac{ 12}{20 }$ = 0.6$

To solve for theta, take the inverse of cosine of both sides:

$\theta + 1 = \cos ^ {-1} (0.6)$

$\theta \approx $53.13^o$$ according to the calculator. That's just one potential value, though. The other angle that would have a cosine of positive 0.6 would be 53.13 degrees below the x-axis in quadrant IV, so subtract from 360:

That gives us two values for $\theta + 1$ , so to get theta we have to subtract 1:

$\theta = $52.13^o$ , $305.87^o$$

Solve for $\theta$ : $3 \cos ^2 \theta - 5 \cos \theta - 2 = 0$

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First solve for $\cos \theta$ using the quadratic formula:

$\cos \theta = $\frac{ 5 \pm \sqrt{25 - 4(3)(-2)}$}{6} = $\frac{5 \pm 7 }{6}$$

One answer is $\cos \theta = $\frac{ 5 + 7 }{6 }$ = $\frac{ 12}$ {6 } = 2$ this is outside the range for cosine, so it does not work as a solution

The other answer is $\cos \theta = $\frac{ 5 - 7 }{6 }$ = $\frac{ -2}{6 }$ = $\frac{ -1}{3 }$$

To solve for theta, take the inverse cosine using a calculator:

$\theta = \cos ^ {-1} (-$\frac{1}{3}$ ) \approx 109.47$

This is just one answer for theta, in quadrant II. Cosine is also negative in quadrant III, so we want to find the angle there with the same cosine. This would be , or

Which is not a solution for $\theta$ : $\tan ^2 (3 \theta ) - 5 \tan (3 \theta ) - 14 = 0$

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To solve, use the quadratic formula:

$\tan ( 3 \theta ) = $\frac{ 5 \pm $\sqrt{5^2$ - 4(1)(-14)}$}{2(1)} = $\frac{ 5 \pm 9 }{ 2 }$$

This gives two solutions.

The first is:

$\tan ( 3 \theta ) = $\frac{ 5 + 9 }{ 2 }$ = $\frac{ 14}{2}$ = 7$

Using a calculator gives us $3 \theta = \tan ^ {-1} (7 ) \approx 81.87$

This is just one potential value, the one in quadrant I. Tangent is also positive in quadrant III, and we can get this angle by adding 180:

The second solution from the quadratic formula is:

$\tan ( 3 \theta ) = $\frac{ 5 - 9 }{2 }$ = $\frac{ -4 }{2}$ = -2$

Using a calculator gives us $3 \theta = \tan ^{-1 } (-2 ) \approx -63.43$ , which we can add to 360 to get as a positive value, .

This is just one potential value, the one in quadrant IV. Tangent is also negative in quadrant II, and we can get this angle by subtracting 180:

Dividing all four of these angles by 3 gives us $\theta = $27.29^o$, $38.86^o$, $87.29^o$, $98.86^o$$

Solve for $\theta$ :

$y = 2 \cos ^2 (3 \theta ) + 5 \cos ( 3 \theta ) - 3$

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Use the quadratic formula to solve for $\cos (3 \theta )$ :

$\cos (3 \theta ) = $\frac{ -5 \pm \sqrt{25 - 4(2)(-3)}$}{2(2)} = $\frac{ -5 \pm \sqrt{49}$}{4} = $\frac{ -5 \pm 7 }{4}$$

One possible solution is:

$\cos (3 \theta ) = $\frac{- 5 - 7 }{4}$ = $\frac{-12}{4}$ = -3$ this is outside of the possible range for cosine

The other solution is:

$\cos (3 \theta ) = $\frac{ -5 + 7 }{4}$ = $\frac{ 2}{4}$ =\frac{ 1}{2}$$

$3 \theta = \cos ^ {-1 } ($\frac{1}{2}$ ) = $\frac{ \pi }{3}$ , $\frac{ 5 \pi }{3}$$ divide by 3

$\theta = $\frac{ \pi }{ 9 }$ , $\frac{ 5 \pi }{ 9 }$$

Solve for $\theta$ :

$0 = 4 \sin ^ 4 \theta + 9 \sin ^ 2 \theta - 9$

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Solve using the quadratic formula:

$\sin ^ 2 \theta = $\frac{ -9 \pm \sqrt{ 81 - 4(4)(-9)}$}{2(4)} = $\frac{ -9 \pm \sqrt{225}$}{8 } = $\frac{ -9 \pm 15}{8}$$

One possible answer is:

$\sin ^ 2 \theta = $\frac{ - 9 + 15 }{ 8 }$ = $\frac{ 6 }{ 8 }$ = $\frac{ 3}{4}$$ take the square root

$\sin \theta = \pm $\sqrt{ \frac{ 3}{4}$ } = \pm $\frac{ \sqrt3}{2}$$

$\theta = \sin ^ {-1} ( \pm $\frac{ \sqrt3}{2}$ )$

$\theta = $\frac{ \pi }{3}$ , $\frac{2 \pi }{3}$ , $\frac{ 4 \pi }{3}$ , $\frac{ 5 \pi }{3}$$

The other would be:

$\sin ^ 2 \theta = $\frac{ -9 - 15 }{8 }$ = $\frac{ - 24}{8 }$ = -3$ this is outside of the range for sine

Solve for $\theta$ : $5 = 2 \cos ^2 \theta - 5 \cos \theta + 2$

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Subtracting 5 from both sides gives the quadratic equation $0 = 2 \cos ^2 \theta - 5 \cos \theta - 3$

Using the quadratic formula gives:

$\cos \theta = $\frac{ 5 \pm \sqrt{25 - 4(2)(-3)}$}{2(2)} = $\frac{ 5 \pm 7}{4}$ = $\frac{ -1}{2}$ , 3$

The cosine cannot be 3 because that's greater than 1.

$\cos \theta = - $\frac{1}{2}$$

$\theta = \cos ^{-1} (-$\frac{1}{2}$) = $\frac{2 \pi }{3}$ , $\frac{ 4 \pi }{3}$$