What is the general solution for 4cos^2(x)+9sin(x)-9=0 ?

The general solution for 4cos^2(x)+9sin(x)-9=0 is x= pi/2+2pin

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4cos^2(x)+9sin(x)-9=0

4 cos^{2} (x) + 9 sin (x) - 9 = 0

x = \frac{π}{2} + 2 πn

Solution steps

4 cos^{2} (x) + 9 sin (x) - 9 = 0

Rewrite using trig identities

- 5 - 4 sin^{2} (x) + 9 sin (x) = 0

Solve by substitution

sin (x) = 1, sin (x) = \frac{5}{4}

sin (x) = 1 : x = \frac{π}{2} + 2 πn

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

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

tan (θ) = - \frac{12}{5}, \frac{π}{2} \leq θ \leq π

cos (x) = sin (2 x + \frac{π}{2}), 0 \leq x \leq 2 π

cos (x) = - 0.397

sin (2 x) cos (x) + cos (2 x) sin (x) = \frac{3}{2}

cos (x) = \frac{5 2}{10}

What is the general solution for 4cos^2(x)+9sin(x)-9=0 ?
The general solution for 4cos^2(x)+9sin(x)-9=0 is x= pi/2+2pin