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

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

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

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

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

Solution steps

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

Subtract

- 5 sin (x)

from both sides

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

cos^{2} (θ) - \frac{1}{4} = 0

2 cot (x) + sec^{2} (x) = 0

π cos (θ) = - \frac{π}{2}

x, y = 1 + sin (\frac{π x}{4})

sin (x) + sin (\frac{x}{2}) = 0

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