# How do I solve 1-sinx=3cosxfor [0, 2)? Thank you!

## $1 - \sin x = 3 \cos x$ for $\left[0 , 2 \pi\right)$

Feb 9, 2018

$x = \frac{\pi}{2} , - 0.9273 . . + 2 \pi$

#### Explanation:

$1 - \sin x = 3 \cos x$

${\left(1 - \sin x\right)}^{2} = {\left(3 \cos x\right)}^{2}$

$1 - 2 \sin x + {\sin}^{2} x = 9 {\cos}^{2} x$

$1 - 2 \sin x + {\sin}^{2} x = 9 \left(1 - {\sin}^{2} x\right)$

$1 - 2 \sin x + {\sin}^{2} x = 9 - 9 {\sin}^{2} x$

$- 8 - 2 \sin x + 10 {\sin}^{2} x = 0$

Let $u = \sin x$:

$10 {u}^{2} - 2 u - 8 = 0$

$5 {u}^{2} - u - 4 = 0$

$\left(u - 1\right) \left(5 u + 4\right) = 0$

$u = 1 , - \frac{4}{5}$

Plug $\sin x$ back in for $u$:

$\sin x = 1 , - \frac{4}{5}$

$x = \frac{\pi}{2} , {\sin}^{-} 1 \left(- \frac{4}{5}\right) \approx - 0.9273 . .$

To get the second number in the wanted range, just add $2 \pi$:

$x = \frac{\pi}{2} , - 0.9273 . . + 2 \pi$