Question

4)
Solve the following equation for exact solution
sin 2x + sin x + 2 cos x + 1 = 0

Answer 1

`\displaystyle\\\\Answer:\ \\x=(3)/(2) \pi +2\pi \mathbb N \\\\x=(2)/(3)\pi +2\pi \mathbb N\\\\x=(4)/(3) \pi +2\pi \mathbb N`

Explanation:

`\displaystyle\\sin(2x)+sin(x)+2cos(x)+1=0\\\\2sin(x)cos(x)+sin(x)+2cos(x)+1=0\\\\2sin(x)cos(x)+2cos(x)+sin(x)+1=0\\\\2cos(x)(sin(x)+1)+(sin(x)+1)=0\\\\(sin(x)+1)(2cos(x)+1)=0\\\\a)\ sin(x)+1=0\\\\sin(x)=-1\\\\x=(3)/(2) \pi +2\pi \mathbb N\\\\b)\ 2cos(x)+1=0\\\\2cos(x)=-1`

Divide both parts of the equation by 2:

`\displaystyle\\cos(x)=-(1)/(2) \\\\x=(2)/(3)\pi +2\pi \mathbb N\\\\x=(4)/(3) \pi +2\pi \mathbb N`