What is the general solution for 9-12sin(x)=4cos^2(x) ?

The general solution for 9-12sin(x)=4cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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9-12sin(x)=4cos^2(x)

9 - 12 sin (x) = 4 cos^{2} (x)

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

Solution steps

9 - 12 sin (x) = 4 cos^{2} (x)

Subtract

4 cos^{2} (x)

from both sides

9 - 12 sin (x) - 4 cos^{2} (x) = 0

Rewrite using trig identities

5 - 12 sin (x) + 4 sin^{2} (x) = 0

Solve by substitution

sin (x) = \frac{5}{2}, sin (x) = \frac{1}{2}

sin (x) = \frac{5}{2} :

No Solution

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

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

5 sec (x) + 7 = - 3

cos (2 θ) + 2 cos (θ) = - 1

(\frac{tan ( x )}{1})^{- 1} = \frac{20}{9.7}

1 - tan (x) = sec (x)

What is the general solution for 9-12sin(x)=4cos^2(x) ?
The general solution for 9-12sin(x)=4cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin