Question

Solve the following equations for x where 0 ≤ x < 2π.
tan(x)sin(2x) = √3 sin (x)
and
cos(2x)−cos(x) = −sin2(x)+1/4

Stuck on these particular types. Have no idea how to start.

Answer 1

tan(x)sin(2x) = √3 sin (x)
tanx(2sinxcosx) = √3sinx
divide by sinx
2tanxcosx = √3
2(sinx/cosx)(cosx) = √3
sinx = √3/2
I recognize x as one of the special angles, x = 60°
but sine is also positive in quadrant II
so x = 180-60 = 120°

x = 60° or x = 120°

for the second, I am confused by your notation.
at the front you have cos(2x) , no problem there, but in
sin2(x) , did you mean sin(2x) or sin^2 x ?

Answer 2

Apologies, I meant" -((sin^2)x + 1/4) "