What is the general solution for 4sin^2(x)+4cos(x)-5=0 ?

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

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

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

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

Rewrite using trig identities

- 1 + 4 cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

2 tan (θ) = sin (θ)

csc (θ) = - 1

tan (θ) = - \frac{1}{3}

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

sin (x) = 0.9

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