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Precalculus : Solve Trigonometric Equations and Inequalities in Quadratic Form

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Example Questions

Example Question #3 : Solving Trigonometric Equations And Inequalities

If $\theta$ $\displaystyle \theta$ exists in the domain from $\left[0,\frac{\pi}{2}\right]$ $\displaystyle \left[0,\frac{\pi}{2}\right]$, solve the following: $0=1-\cos^2\theta$ $\displaystyle 0=1-\cos^2\theta$

Possible Answers:

$\displaystyle 0$

$\textup{No solution.}$ $\displaystyle \textup{No solution.}$

$\pi$ $\displaystyle \pi$

$0,\frac{\pi}{2}$ $\displaystyle 0,\frac{\pi}{2}$

$0,\pi$ $\displaystyle 0,\pi$

Correct answer:

$\displaystyle 0$

Explanation:

Factorize $0=1-cos^2\theta$ $\displaystyle 0=1-cos^2\theta$.

$1-cos^2\theta= (1+cos(\theta))(1-cos(\theta))$ $\displaystyle 1-cos^2\theta= (1+cos(\theta))(1-cos(\theta))$

Set both terms equal to zero and solve.

$1+cos(\theta)=0$ $\displaystyle 1+cos(\theta)=0$

$cos(\theta)=-1$ $\displaystyle cos(\theta)=-1$

$\theta=\pi$ $\displaystyle \theta=\pi$

This value is not within the $[0,\frac{\pi}{2}]$ $\displaystyle [0,\frac{\pi}{2}]$ domain.

$1-cos(\theta)=0$ $\displaystyle 1-cos(\theta)=0$

$cos(\theta)=1$ $\displaystyle cos(\theta)=1$

$\theta=0$ $\displaystyle \theta=0$

This is the only correct value in the $[0,\frac{\pi}{2}]$ $\displaystyle [0,\frac{\pi}{2}]$ domain.

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Example Question #13 : Trigonometric Identities

Solve for $\theta$ $\displaystyle \theta$ in the equation $sin^2(\theta)-2sin(\theta)+1=0$ $\displaystyle sin^2(\theta)-2sin(\theta)+1=0$ on the interval $0\leq\theta\leq2\pi$ $\displaystyle 0\leq\theta\leq2\pi$.

Possible Answers:

$\theta=\frac{3\pi}{2}$ $\displaystyle \theta=\frac{3\pi}{2}$

$\theta=\frac{3\pi}{4}$ $\displaystyle \theta=\frac{3\pi}{4}$

$\theta=\frac{\pi}{2}$ $\displaystyle \theta=\frac{\pi}{2}$

$\theta=0$ $\displaystyle \theta=0$

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

Correct answer:

$\theta=\frac{\pi}{2}$ $\displaystyle \theta=\frac{\pi}{2}$

Explanation:

If you substitute $x=sin(\theta)$ $\displaystyle x=sin(\theta)$ you obtain a recognizable quadratic equation which can be solved for $\displaystyle x$,

$\displaystyle (x-1)^2=0$.

Then we can plug $\displaystyle x$ back into our equation and use the unit circle to find that

$\theta=\frac{\pi}{2}$ $\displaystyle \theta=\frac{\pi}{2}$.

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Example Question #4 : Solving Trigonometric Equations And Inequalities

Given that theta exists from $[0,2\pi]$ $\displaystyle [0,2\pi]$, solve: $2cos^2\theta=cos(\theta)$ $\displaystyle 2cos^2\theta=cos(\theta)$

Possible Answers:

$\frac{\pi}{2},\frac{3\pi}{2}$ $\displaystyle \frac{\pi}{2},\frac{3\pi}{2}$

$\frac{\pi}{2}$ $\displaystyle \frac{\pi}{2}$

$\frac{\pi}{3}, \frac{5\pi}{3}$ $\displaystyle \frac{\pi}{3}, \frac{5\pi}{3}$

$\textup{None of the given answers.}$ $\displaystyle \textup{None of the given answers.}$

$\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{3}$ $\displaystyle \frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{3}$

Correct answer:

$\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{3}$ $\displaystyle \frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{3}$

Explanation:

In order to solve $2cos^2\theta=cos(\theta)$ $\displaystyle 2cos^2\theta=cos(\theta)$ appropriately, do not divide $cos(\theta)$ $\displaystyle cos(\theta)$ on both sides. The effect will eliminate one of the roots of this trig function.

Substract $cos(\theta)$ $\displaystyle cos(\theta)$ from both sides.

$2cos^2\theta-cos(\theta)=0$ $\displaystyle 2cos^2\theta-cos(\theta)=0$

Factor the left side of the equation.

$cos(\theta)[2cos(\theta)-1]=0$ $\displaystyle cos(\theta)[2cos(\theta)-1]=0$

Set each term equal to zero, and solve for theta with the restriction $[0,2\pi]$ $\displaystyle [0,2\pi]$.

$cos(\theta)=0$ $\displaystyle cos(\theta)=0$

$\theta = \frac{\pi}{2},\frac{3\pi}{2}$ $\displaystyle \theta = \frac{\pi}{2},\frac{3\pi}{2}$

$2cos(\theta)-1=0$ $\displaystyle 2cos(\theta)-1=0$

$2cos(\theta)=1$ $\displaystyle 2cos(\theta)=1$

$cos(\theta)=\frac{1}{2}$ $\displaystyle cos(\theta)=\frac{1}{2}$

$\theta= \frac{\pi}{3}, \frac{5\pi}{3}$ $\displaystyle \theta= \frac{\pi}{3}, \frac{5\pi}{3}$

The correct answer is:

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

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Example Question #1 : Solving Trigonometric Equations And Inequalities

Solve $\sin^2(x) + 4 = 0$ $\displaystyle \sin^2(x) + 4 = 0$ for $0^{\circ}\leq x\leq 360^{\circ}$ $\displaystyle 0^{\circ}\leq x\leq 360^{\circ}$

Possible Answers:

x = 0 $\displaystyle x = 0$

There is no solution.

$\displaystyle x = -2, x = 2$

$x = 90^{\circ}$ $\displaystyle x = 90^{\circ}$

Correct answer:

There is no solution.

Explanation:

By subtracting $\displaystyle 4$ from both sides of the original equation, we get $\sin^2(x) = -4$ $\displaystyle \sin^2(x) = -4$. We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.

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Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Solve $\sin^2(x)-4 = 0$ $\displaystyle \sin^2(x)-4 = 0$ when $0^{\circ}\leq x \leq 360^{\circ}$ $\displaystyle 0^{\circ}\leq x \leq 360^{\circ}$

Possible Answers:

x= 0 $\displaystyle x= 0$

There are no solutions.

$x = 45^{\circ}$ $\displaystyle x = 45^{\circ}$

$\displaystyle x=2, x=-2$

Correct answer:

There are no solutions.

Explanation:

Given that, for any input, $-1\leq \sin(x) \leq1$ $\displaystyle -1\leq \sin(x) \leq1$, we know that $-1\leq\sin^2(x)\leq1$ $\displaystyle -1\leq\sin^2(x)\leq1$, and so the equation $\sin^2(x)=4$ $\displaystyle \sin^2(x)=4$ can have no solutions.

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Example Question #2 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Solve $\sin^2(x)-1 =0$ $\displaystyle \sin^2(x)-1 =0$ when $0^{\circ}\leq x \leq 360^{\circ}$ $\displaystyle 0^{\circ}\leq x \leq 360^{\circ}$

Possible Answers:

$x = 90^{\circ}, x=270^{\circ}$ $\displaystyle x = 90^{\circ}, x=270^{\circ}$

$x = 90^{\circ}$ $\displaystyle x = 90^{\circ}$

There are no solutions.

$x = 0^{\circ}$ $\displaystyle x = 0^{\circ}$

Correct answer:

$x = 90^{\circ}, x=270^{\circ}$ $\displaystyle x = 90^{\circ}, x=270^{\circ}$

Explanation:

By adding one to both sides of the original equation, we get $\sin^2(x) = 1$ $\displaystyle \sin^2(x) = 1$, and by taking the square root of both sides of this, we get $\sin(x) = 1, \sin(x) = -1.$ $\displaystyle \sin(x) = 1, \sin(x) = -1.$ From there, we get that, on the given interval, the only solutions are $x = 90^{\circ}$ $\displaystyle x = 90^{\circ}$ and $x = 270^{\circ}$ $\displaystyle x = 270^{\circ}$.

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Precalculus : Solve Trigonometric Equations and Inequalities in Quadratic Form

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Solving Trigonometric Equations And Inequalities

Example Question #13 : Trigonometric Identities

Example Question #4 : Solving Trigonometric Equations And Inequalities

Example Question #1 : Solving Trigonometric Equations And Inequalities

Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Example Question #2 : Solve Trigonometric Equations And Inequalities In Quadratic Form

All Precalculus Resources

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