Question

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

Help solve, please.
solve `"sin"^4 "x" -5"cos"^2 "x" +1=0`

Answer 1

`56^@93; 123^@07; 236^@93; 303^@07` --> + `(k360^@)`

Explanation:

`sin^4x - 5(1 - sin^2 x) + 1 = 0`
`sin^4 x - 5 + 5sin^2 x + 1= 0`
`sin^4 x + 5sin^2 x - 4 = 0`
Call `sin^2 x = t`, we get a quadratic equation in t.
`t^2 + 5t - 4 = 0`
`D = d^2 = b^2 - 4ac = 25 + 16 = 41` --> `d= +- sqrt41`
There are 2 real roots:
`t = - 5/2 +- sqrt41/2 = (-5 +- sqrt41)/2`
a. `t = sin^2 x = (- 5 + sqrt41)/2 = 0.70`
`sin x = +- 0.838`
Unit circle and calculator give -->
1. `sin x = 0.838` -->
`x = 56^@93`, and `x = 180 - 56^@93 = 123^@07`
2. sin x = - 0.838 -->
`x = - 56^@93`, or `x = 360 - 56.93 = 303^@07` (co-terminal), and
`x = 180 - (- 56.93) = 180 + 56.93 = 236^@93`
b. `sin^2 x = t = - 5/2 - sqrt41/2 = (-5 - sqrt41)/2`. (Rejected)
For general answers, add `k360^@`