Question

1) 3cos x -\(\sqrt{3}\) sin x = -3 

2) sin 4x = \(\sqrt{3}\)(cos 4x -1)

Answer 1

`1)3cos x-\sqrt{3}sin x=-3`

`<=>\sqrt{3}/2cos x-1/2sin x=-\sqrt{3}/2`

`<=>cos(x+\pi/6)=-\sqrt{3}/2`

`<=>[(x+\pi/6=[5\pi]/6+k2\pi),(x+\pi/6=[-5\pi]/6+k2\pi):}`

`<=>[(x=[2\pi]/3+k2\pi),(x=-\pi+k2\pi):}`    `(k in ZZ)`

_______________________________________________

`2)sin 4x=\sqrt{3}(cos 4x-1)`

`<=>sin 4x-\sqrt{3}cos 4x=-\sqrt{3}`

`<=>1/2sin 4x-\sqrt{3}/2cos 4x=-\sqrt{3}/2`

`<=>sin(4x-\pi/3)=-\sqrt{3}/2`

`<=>[(4x-\pi/3=-\pi/3+k2\pi),(4x-\pi/3=[4\pi]/3+k2\pi):}`

`<=>[(x=k\pi/2),(x=[5\pi]/12+k\pi/2):}`       `(k in ZZ)`