What is the general solution for 1/(cos^2(x))= 2/(3-3sin(x)) ?

The general solution for 1/(cos^2(x))= 2/(3-3sin(x)) is x= pi/6+2pin,x=(5pi)/6+2pin

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1/(cos^2(x))= 2/(3-3sin(x))

\frac{1}{cos ^{2} ( x )} = \frac{2}{3 - 3 sin ( x )}

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

Solution steps

\frac{1}{cos ^{2} ( x )} = \frac{2}{3 - 3 sin ( x )}

Subtract

\frac{2}{3 - 3 sin ( x )}

from both sides

\frac{1}{cos ^{2} ( x )} - \frac{2}{3 - 3 sin ( x )} = 0

Simplify

\frac{1}{cos ^{2} ( x )} - \frac{2}{3 - 3 sin ( x )} : - \frac{- 3 ( sin ( x ) - 1 ) - 2 cos ^{2} ( x )}{3 cos ^{2} ( x ) ( sin ( x ) - 1 )}

- \frac{- 3 ( sin ( x ) - 1 ) - 2 cos ^{2} ( x )}{3 cos ^{2} ( x ) ( sin ( x ) - 1 )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- (- 3 (sin (x) - 1) - 2 cos^{2} (x)) = 0

Rewrite using trig identities

- 1 - 2 sin^{2} (x) + 3 sin (x) = 0

Solve by substitution

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

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{π}{2} + 2 πn

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

cos (4 x) = - 1

0.94 = 0.33 sin (x) + cos (x)

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

cos (x) = sec (x)

cos^{2} (x) = 2

What is the general solution for 1/(cos^2(x))= 2/(3-3sin(x)) ?
The general solution for 1/(cos^2(x))= 2/(3-3sin(x)) is x= pi/6+2pin,x=(5pi)/6+2pin