What is the general solution for 5sin(x)=2cos^2(x)-4 ?

The general solution for 5sin(x)=2cos^2(x)-4 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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5sin(x)=2cos^2(x)-4

5 sin (x) = 2 cos^{2} (x) - 4

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

Solution steps

5 sin (x) = 2 cos^{2} (x) - 4

Subtract

2 cos^{2} (x) - 4

from both sides

5 sin (x) - 2 cos^{2} (x) + 4 = 0

Rewrite using trig identities

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

Solve by substitution

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

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

sin (x) = - 2 :

No Solution

Combine all the solutions

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

2 sin (x - \frac{π}{2}) + 2 = 0

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

2 cos^{3} (x) - cos (x) = 0

- 2 cos (2 x) = 3

(tan (x) - 1) (2 sin (x) - 3) = 0

What is the general solution for 5sin(x)=2cos^2(x)-4 ?
The general solution for 5sin(x)=2cos^2(x)-4 is x=(7pi)/6+2pin,x=(11pi)/6+2pin