Trigonometry Question can't reach the solution please help.

[seema singh]

Solve the equation sin2x = √3cos (3π/2 - x) and write the roots of this equation that belong

to the interval [ π:7π/2] .
I don't have a clue how to even start solving this equation. Can anyone help me with it?
I am trying to convert sin to cos or cos to sin but can't reach a solution.

 

[Otis]

Hello. Here's a hint: cos(3π/2 - x) = -sin(x).

?

 

[MarkFL]

If we use a double-angle identity for sine on the LHS, and an angle difference identity for cosine on the right, we get:

[MATH]2\sin(x)\cos(x)=-\sqrt{3}\sin(x)[/MATH]
Arrange as:

[MATH]2\sin(x)\cos(x)+\sqrt{3}\sin(x)=0[/MATH]
Factor:

[MATH]\sin(x)(2\cos(x)+\sqrt{3})=0[/MATH]
Can you proceed?