What is the general solution for (2cos^2(x))/(2(1-sin(x))-cos^2(x))=0 ?

The general solution for (2cos^2(x))/(2(1-sin(x))-cos^2(x))=0 is x=(3pi)/2+2pin

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(2cos^2(x))/(2(1-sin(x))-cos^2(x))=0

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

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

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

2 cos^{2} (x) = 0

Divide both sides by

2

2 cos^{2} (x) = 0

Divide both sides by

2

2 cos^{2} (x) = 0

Divide both sides by

2

\frac{2 cos ^{2} ( x )}{2} = \frac{0}{2}

Simplify

cos^{2} (x) = 0

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

2 sin (x) = - 3

tan (θ) = \frac{- 12}{5}, cot (θ)

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

0 = 3 cos (θ)

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

What is the general solution for (2cos^2(x))/(2(1-sin(x))-cos^2(x))=0 ?
The general solution for (2cos^2(x))/(2(1-sin(x))-cos^2(x))=0 is x=(3pi)/2+2pin