Question

How do you solve `-4sin^2x=-3`?

Answer 1

The two solutions are `60°` and `300°`

Explanation:

First we do some cleanup:

`sin^2x = (-3)/(-4) = 0,75`

Next we find the root of `sin^2x`

`sinx = +- 0.866`

We need the `+-` here, because a square is always positive.

So solution (1) is `arcsin(0.866) = 60°`
And solution (2) is `arcsin(-0.866) = 300°`

Note: `arcsin and sin^-1` are equivalent.

Answer 2

Hint: `color(blue)((1)2sin^2theta=1-cos2theta`
`color(blue)((2) costheta=cosalpha=>theta=2kpi+-alpha,kinZ`
Ans.: `color(red)(x=kpi+-pi/3,kinZ`

Explanation:

We note that

`4sin^2x=-3=>sin^2x=-3/4=>(sinx)^2=-3/4<0`

This is not possible.

So,we take

`-4sin^2x=-3=>color(red)(4sin^2x=3...to` (1)

`=>2sin^2x=3/2`

`:.1-cos2x=3/2=>cos2x=-1/2`

`=>cos2x=cos((2pi)/3)`

`2x=2kpi+-(2pi)/3,kinZ`

`=>color(red)(x=kpi+-pi/3,kinZ`