# Solve "sin"^4 "x" -5"cos"^2 "x" +1=0 ?

## Help solve, please. solve $\text{sin"^4 "x" -5"cos"^2 "x} + 1 = 0$

Jun 3, 2018

56^@93; 123^@07; 236^@93; 303^@07 --> + $\left(k {360}^{\circ}\right)$

#### Explanation:

${\sin}^{4} x - 5 \left(1 - {\sin}^{2} x\right) + 1 = 0$
${\sin}^{4} x - 5 + 5 {\sin}^{2} x + 1 = 0$
${\sin}^{4} x + 5 {\sin}^{2} x - 4 = 0$
Call ${\sin}^{2} x = t$, we get a quadratic equation in t.
${t}^{2} + 5 t - 4 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 25 + 16 = 41$ --> $d = \pm \sqrt{41}$
There are 2 real roots:
$t = - \frac{5}{2} \pm \frac{\sqrt{41}}{2} = \frac{- 5 \pm \sqrt{41}}{2}$
a. $t = {\sin}^{2} x = \frac{- 5 + \sqrt{41}}{2} = 0.70$
$\sin x = \pm 0.838$
Unit circle and calculator give -->
1. $\sin x = 0.838$ -->
$x = {56}^{\circ} 93$, and $x = 180 - {56}^{\circ} 93 = {123}^{\circ} 07$
2. sin x = - 0.838 -->
$x = - {56}^{\circ} 93$, or $x = 360 - 56.93 = {303}^{\circ} 07$ (co-terminal), and
$x = 180 - \left(- 56.93\right) = 180 + 56.93 = {236}^{\circ} 93$
b. ${\sin}^{2} x = t = - \frac{5}{2} - \frac{\sqrt{41}}{2} = \frac{- 5 - \sqrt{41}}{2}$. (Rejected)
For general answers, add $k {360}^{\circ}$