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

The general solution for cos(2θ)+sin(θ)+2=0 is θ=(3pi)/2+2pin

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cos(2θ)+sin(θ)+2=0

cos (2 θ) + sin (θ) + 2 = 0

θ = \frac{3 π}{2} + 2 πn

Solution steps

cos (2 θ) + sin (θ) + 2 = 0

Rewrite using trig identities

3 + sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = \frac{3}{2}

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = \frac{3}{2} :

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

cos (x) = \frac{10}{50}

- 7 tan (x) + 3 = - 3 (3 + tan (x))

csc^{2} (x) + 3 cot^{2} (x) = 1

2 sin^{2} (θ) - sin (θ) - 3 = 0

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

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