What is the general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi ?

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

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sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{3 π}{2}

Solution steps

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

Rewrite using trig identities

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

Factor

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

(1 + 2 cos (θ)) (sin (θ) + 1) = 0

Solving each part separately

1 + 2 cos (θ) = 0 or sin (θ) + 1 = 0

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

sin (θ) + 1 = 0, 0 \leq θ \leq 2 π : θ = \frac{3 π}{2}

Combine all the solutions

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{3 π}{2}

sin^{2} (t) = cos^{2} (t)

cos (x) - sin (x) sin (2 x) = 0

4 sin (x) + 3 cos (x) = 4

sin (θ) - cos (θ) = 0, 0 \leq θ \leq 2 π

3 sec^{2} (θ) - 12 = 5 sec (θ)

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