Solve Trigonometric Equations and Inequalities in Quadratic Form - Pre-Calculus

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Solve for $\theta$ in the equation on the interval $0\leq\theta\leq2\pi$ .

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

.

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

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

If $\theta$ exists in the domain from $\left[0,$\frac{\pi}{2}\right]$ , solve the following:

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Factorize .

Set both terms equal to zero and solve.

$1+cos(\theta)=0$

$cos(\theta)=-1$

$\theta=\pi$

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

$1-cos(\theta)=0$

$cos(\theta)=1$

$\theta=0$

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

Given that theta exists from $[0,2\pi]$ , solve:

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

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

Factor the left side of the equation.

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

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

$cos(\theta)=0$

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

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

$2cos(\theta)=1$

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

$\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}$$

Solve for

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By subtracting from both sides of the original equation, we get . We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.

Solve when

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

Solve when

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

Solve for $\theta$ in the equation on the interval $0\leq\theta\leq2\pi$ .

Tap to see back →

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

.

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

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

If $\theta$ exists in the domain from $\left[0,$\frac{\pi}{2}\right]$ , solve the following:

Tap to see back →

Factorize .

Set both terms equal to zero and solve.

$1+cos(\theta)=0$

$cos(\theta)=-1$

$\theta=\pi$

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

$1-cos(\theta)=0$

$cos(\theta)=1$

$\theta=0$

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

Given that theta exists from $[0,2\pi]$ , solve:

Tap to see back →

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

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

Factor the left side of the equation.

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

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

$cos(\theta)=0$

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

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

$2cos(\theta)=1$

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

$\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}$$

Solve for

Tap to see back →

By subtracting from both sides of the original equation, we get . We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.

Solve when

Tap to see back →

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

Solve when

Tap to see back →

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

Solve for $\theta$ in the equation on the interval $0\leq\theta\leq2\pi$ .

Tap to see back →

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

.

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

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

If $\theta$ exists in the domain from $\left[0,$\frac{\pi}{2}\right]$ , solve the following:

Tap to see back →

Factorize .

Set both terms equal to zero and solve.

$1+cos(\theta)=0$

$cos(\theta)=-1$

$\theta=\pi$

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

$1-cos(\theta)=0$

$cos(\theta)=1$

$\theta=0$

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

Given that theta exists from $[0,2\pi]$ , solve:

Tap to see back →

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

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

Factor the left side of the equation.

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

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

$cos(\theta)=0$

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

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

$2cos(\theta)=1$

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

$\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}$$

Solve for

Tap to see back →

By subtracting from both sides of the original equation, we get . We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.

Solve when

Tap to see back →

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

Solve when

Tap to see back →

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

Solve for $\theta$ in the equation on the interval $0\leq\theta\leq2\pi$ .

Tap to see back →

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

.

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

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

If $\theta$ exists in the domain from $\left[0,$\frac{\pi}{2}\right]$ , solve the following:

Tap to see back →

Factorize .

Set both terms equal to zero and solve.

$1+cos(\theta)=0$

$cos(\theta)=-1$

$\theta=\pi$

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

$1-cos(\theta)=0$

$cos(\theta)=1$

$\theta=0$

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