What is the general solution for 4sin(x)-13=2cos^2(x)-9 ?

The general solution for 4sin(x)-13=2cos^2(x)-9 is x= pi/2+2pin

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4sin(x)-13=2cos^2(x)-9

4 sin (x) - 13 = 2 cos^{2} (x) - 9

x = \frac{π}{2} + 2 πn

Solution steps

4 sin (x) - 13 = 2 cos^{2} (x) - 9

Subtract

2 cos^{2} (x) - 9

from both sides

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

Rewrite using trig identities

- 6 + 2 sin^{2} (x) + 4 sin (x) = 0

Solve by substitution

sin (x) = 1, sin (x) = - 3

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = - 3 :

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

cos (θ) = - \frac{8}{11}

sin (θ) = \frac{3}{12}

cos (x) = \frac{12}{17}

x, (d^{3} + d^{2} + 1) y = cos (3 x)

arccos (2 x - 1) = \frac{π}{6}

What is the general solution for 4sin(x)-13=2cos^2(x)-9 ?
The general solution for 4sin(x)-13=2cos^2(x)-9 is x= pi/2+2pin