What is the general solution for 2cos(2x)+1+4sin(x)=0 ?

The general solution for 2cos(2x)+1+4sin(x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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2cos(2x)+1+4sin(x)=0

2 cos (2 x) + 1 + 4 sin (x) = 0

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

Solution steps

2 cos (2 x) + 1 + 4 sin (x) = 0

Rewrite using trig identities

3 + 4 sin (x) - 4 sin^{2} (x) = 0

Solve by substitution

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

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

sin (x) = \frac{3}{2} :

No Solution

Combine all the solutions

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

tan (x) = - 1, tan (π + x) = y

sin (θ) sec (θ) - sin (θ) = 0

r = 2 a cos (x)

tan (t) = - \frac{1}{3}, - π < t \leq π

sin (x) = \frac{- 4}{5}

What is the general solution for 2cos(2x)+1+4sin(x)=0 ?
The general solution for 2cos(2x)+1+4sin(x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin