Question

Solve the equation `5sin4x=2sin2x` if `0^@ <= x <= 360^@`?

Answer 1

Solution is `{0^o,39.232^o,90^o,140.768^o,180^o,219.232^o,270^o,320.768^o,360^o}`

Explanation:

`5sin4x=2sin2x`

`hArr5xx2sin2xcos2x=2sin2x`

or `10sin2xcos2x-2sin2x=0`

or `2sin2x(5cos2x-1)=0`

Hence either `sin2x=0` i.e. `2x={0^o,180^o,360^o,540^o,720^o,....}`

or in the given range `0<=x<=360^o`, `x={0^o,90^o,180^o,270^o,360^o}`

or `5cos2x-1=0` i.e. `cos2x=1/5=cos(+-78.463^o)` i.e.

`2x={78.463^o,(360^o-78.463^o),(360^o +78.463^o),(720^o-78.463^o)}`

`x={39.232^o,140.768^o,219.232^o,320.768^o}`

Hence, solution is `{0^o,39.232^o,90^o,140.768^o,180^o,219.232^o,270^o,320.768^o,360^o}`
graph{5sin(4x)-2sin(2x) [-1, 7.26, -4, 6]}