Question

Find the exact solutions of the equation in the intervals [0,2pi)(Enter your answers as a comma separated list)4 sin 2x sin X = 4 cos xX= ?

Answer 1

Step 1

Subtract 4cosx from both sides

`4\text{sin}2\text{x.sinx}-4\cos x=4\cos x-4\cos x`

Simplify

`4\text{sin}2\text{x.sinx}-4\cos x=0`

Step 2

Rewrite using trigonometric identities

`\begin{gathered} \sin 2x=2sinx.\cos x \\ 4(2\sin x\text{cosx})(\sin x)-4\cos x=0 \\ 8\sin ^2(x)\cos (x)-4\cos (x)=0 \end{gathered}`

Step 3

Factorize

`8\sin ^2(x)\cos (x)-4\cos (x)=4\cos (x)(\sqrt[]{2}\sin (x)+1)(\sqrt[]{2}\sin (x)-1)_{}`
`\begin{gathered} 4\cos \mleft(x\mright)\mleft(√(2)\sin \mleft(x\mright)+1\mright)\mleft(√(2)\sin \mleft(x\mright)-1\mright)=0 \\ \end{gathered}`

Solve separately for x

`\begin{gathered} 4\cos x=0 \\ x=\cos ^(-1)(0) \\ x=90\text{ and }270\text{ within the range \lbrack{}0,2}\pi) \\ In\text{ radians }90^o\text{ = }(\pi)/(2),270^o=(3\pi)/(2) \\ \\ \end{gathered}`