Question

Trigonometry Question: Find all solutions on the interval [0,2π) 12sin^2(x)+cos(x)-6=0

Answer 1

Recall the Pythagorean identity,

$\sin^2(x) + \cos^2(x) = 1$

Use it to rewrite the equation in terms of cos only.

$12 (1 - \cos^2(x)) + \cos(x) - 6 = 0$

$-12 \cos^2(x) + \cos(x) + 6 = 0$

Factorize the left side.

$-(3\cos(x) + 2) (4\cos(x) - 3) = 0$

Solve the two cases for $\cos(x)$.

$3 \cos(x) + 2 = 0 \text{ or } 4\cos(x) - 3 = 0$

$\cos(x) = -\dfrac23 \text{ or } \cos(x) = \dfrac34$

From the first equation, we get one family of solutions:

$\cos(x) = -\dfrac23$

$x = \cos^{-1}\left(-\dfrac23\right) + 2n\pi \text{ or } x = -\cos^{-1}\left(-\dfrac23\right) + 2n\pi$

From the second equation, we get another family:

$\cos(x) = \dfrac34$

$x = \cos^{-1}\left(\dfrac34\right) + 2n\pi \text{ or } x = -\cos^{-1}\left(\dfrac34\right) + 2n\pi$

where $n$ is an integer.

We get solutions in the given domain for

$n = 0 \implies \boxed{x = \cos^{-1}\left(-\dfrac23\right)} \text{ or } \boxed{x = \cos^{-1}\left(\dfrac34\right)}$

$n=1 \implies \boxed{x = 2\pi - \cos^{-1}\left(-\dfrac23\right)} \text{ or } \boxed{x = 2\pi - \cos^{-1}\left(\dfrac34\right)}$