Question

How do you solve `sin^2x+ 3/2cos(2x) = 0`?

Answer 1

Use relevant identities.

Explanation:

`sin^2(x) + 3/2cos(2x) = 0` = `sin^2(x) + 3/2(1-2sin^2(x))= 0`
Simplifying will get you
`-2sin^2(x) = -3/2`
`sin^2(x) = 3/4`
`sin(x) = sqrt(3)/2` and `sin(x) = -sqrt(3)/2`
Basic Angle = `sin^-1(sqrt(3)/2)`
Basic Angle = `60^.`
Since `sin` is both positive and negative, angle must lie in all quadrants.
Thus `angle` = `60, 180^(.)-60^(.), 180^(.) + 60^(.), 360^(.) - 60^(.)`