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

2 Answers
Mar 22, 2018

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.

Mar 22, 2018

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