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

The general solution for 4cos^2(x)+4sin(x)-5=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

\frac{3}{3} = - \frac{1}{3} \cdot tan (θ)

tan (x + \frac{π}{3}) = 3

16 sin^{2} (x) = 1

x, - 1 \cdot \frac{cos ( x )}{sin ( x )} = 0

What is the general solution for 4cos^2(x)+4sin(x)-5=0 ?
The general solution for 4cos^2(x)+4sin(x)-5=0 is x= pi/6+2pin,x=(5pi)/6+2pin