What is the general solution for 2sin^2(x)+5cos(x)+1=0 ?

The general solution for 2sin^2(x)+5cos(x)+1=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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2sin^2(x)+5cos(x)+1=0

2 sin^{2} (x) + 5 cos (x) + 1 = 0

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solution steps

2 sin^{2} (x) + 5 cos (x) + 1 = 0

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = 3 :

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

5 sin (5 x) = 0

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

- 7 sin^{2} (θ) + 2 sin (θ) + 7 = 0

cos (x) - 2 sin^{2} (x) + 1 = 0

What is the general solution for 2sin^2(x)+5cos(x)+1=0 ?
The general solution for 2sin^2(x)+5cos(x)+1=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin