What is the general solution for -sin(θ)-cos(2θ)=0 ?

The general solution for -sin(θ)-cos(2θ)=0 is θ= pi/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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-sin(θ)-cos(2θ)=0

- sin (θ) - cos (2 θ) = 0

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

- sin (θ) - cos (2 θ) = 0

Rewrite using trig identities

- 1 - sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

6 cos^{2} (x) + cos (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

tan (y) = \frac{π}{4}

2 sin (x - 1) + 1 = 0

- 6 sin^{2} (x) + sin (x) = - 3

csc (x) + 2 = 0, 0 \leq x \leq 2 π

What is the general solution for -sin(θ)-cos(2θ)=0 ?
The general solution for -sin(θ)-cos(2θ)=0 is θ= pi/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin