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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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.

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Question

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

Answer

In order to solve $2cos^2\theta=cos(\theta)$ 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.

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

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:

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

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Question

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

Answer

By subtracting from both sides of the original equation, we get $\sin^2(x) = -4$ . 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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Question

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

Answer

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

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Question

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

Answer

By adding one to both sides of the original equation, we get $\sin^2(x) = 1$ , 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 $x = 270^{\circ}$ .

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