Question

How to Solve `5-4cosx=4sin^2x` for `-90°<=x<=90°` ?

`5-4cosx=4sin^2x` for `-90°<=x<=90°`

Answer 1

`x=60^@,-60^@`

Explanation:

Identity Required:
`sin^2 x+cos^2 x=1`
`sin^2 x=1-cos^2 x`

Now solving:
`5-4cosx=4sin^2 x`
`5-4cosx=4(1-cos^2 x)`
`5-4cosx=4-4cos^2 x`
`4cos^2 x -4cos x+1=0`
`(2cosx -1)^2=0`
`cosx=1/2`
`x=60^@,-60^@`