What is the general solution for sin(2x)+sin(x)+2cos(x)+1=0 ?

The general solution for sin(2x)+sin(x)+2cos(x)+1=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=(3pi)/2+2pin

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sin(2x)+sin(x)+2cos(x)+1=0

sin (2 x) + sin (x) + 2 cos (x) + 1 = 0

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

sin (2 x) + sin (x) + 2 cos (x) + 1 = 0

Rewrite using trig identities

1 + sin (x) + 2 cos (x) + 2 cos (x) sin (x) = 0

Factor

1 + sin (x) + 2 cos (x) + 2 cos (x) sin (x) : (1 + 2 cos (x)) (sin (x) + 1)

(1 + 2 cos (x)) (sin (x) + 1) = 0

Solving each part separately

1 + 2 cos (x) = 0 or sin (x) + 1 = 0

1 + 2 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sin (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{3 π}{2} + 2 πn

csc (x) = \frac{13}{5}

6 csc (θ) + 7 = 0

cos (2 x) + cos (x) = 0, 0 \leq x \leq 2 π

cos (5 x) cos (3 x) - sin (5 x) sin (3 x) = \frac{3}{2}

- 6 sin (2 x) = 0

What is the general solution for sin(2x)+sin(x)+2cos(x)+1=0 ?
The general solution for sin(2x)+sin(x)+2cos(x)+1=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=(3pi)/2+2pin