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

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

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

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

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

Solution steps

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

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

sin (x) - 2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

cos (x + π) - cos (x) - 1 = 0

sin (x) = \frac{15}{17}

2 cos (x) - sec (x) = 1

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