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) = -\frac23 \text{ or } \cos(x) = \frac34`

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

`\cos(x) = -\frac23`

`x = \cos^(-1)\left(-\frac23\right) + 2n\pi \text{ or } x = -\cos^(-1)\left(-\frac23\right) + 2n\pi`

From the second equation, we get another family:

`\cos(x) = \frac34`

`x = \cos^(-1)\left(\frac34\right) + 2n\pi \text{ or } x = -\cos^(-1)\left(\frac34\right) + 2n\pi`

where
`n` is an integer.

We get solutions in the given domain for

`n = 0 \implies \boxed{x = \cos^(-1)\left(-\frac23\right)} \text{ or } \boxed{x = \cos^(-1)\left(\frac34\right)}`

`n=1 \implies \boxed{x = 2\pi - \cos^(-1)\left(-\frac23\right)} \text{ or } \boxed{x = 2\pi - \cos^(-1)\left(\frac34\right)}`