What is the general solution for cos(2x)+5sin(x)+2=0 ?

The general solution for cos(2x)+5sin(x)+2=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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cos(2x)+5sin(x)+2=0

cos (2 x) + 5 sin (x) + 2 = 0

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

Solution steps

cos (2 x) + 5 sin (x) + 2 = 0

Rewrite using trig identities

3 - 2 sin^{2} (x) + 5 sin (x) = 0

Solve by substitution

sin (x) = - \frac{1}{2}, sin (x) = 3

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

sin (x) = 3 :

No Solution

Combine all the solutions

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

2 sin^{2} (θ) - 3 cos (θ) = 3

cot (θ) = \frac{1}{3}

sec (x) csc (x) = 2 sec (x)

2 tan^{2} (x) + sec (x) = 1

(tan (θ) - 1) (cos (θ) - 1) = 0

What is the general solution for cos(2x)+5sin(x)+2=0 ?
The general solution for cos(2x)+5sin(x)+2=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin