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

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

There is no solution.

$x = 90^{\circ}$

Explanation

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.

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

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

There are no solutions.

$x = 90^{\circ}$

$x = 0^{\circ}$

Explanation

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 $x = 270^{\circ}$ .

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

$\pi$

$0,\pi$

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

$\textup{No solution.}$

Explanation

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.

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

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

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

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

$\frac{\pi}{2}$

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

Explanation

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.