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

The general solution for 1-(cos^2(x))/(1+sin(x))=2cos(x) is x=1.10714…+pin

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

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

x = 1.10714 \dots + πn

Solution steps

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

Subtract

2 cos (x)

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Factor

sin (x) + sin^{2} (x) - (1 + sin (x)) \cdot 2 cos (x) : (1 + sin (x)) (sin (x) - 2 cos (x))

(1 + sin (x)) (sin (x) - 2 cos (x)) = 0

Solving each part separately

1 + sin (x) = 0 or sin (x) - 2 cos (x) = 0

1 + sin (x) = 0 : x = \frac{3 π}{2} + 2 πn

sin (x) - 2 cos (x) = 0 : x = arctan (2) + πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = arctan (2) + πn

Show solutions in decimal form

x = 1.10714 \dots + πn

cos (x) = \frac{4}{14}

\frac{sin ( 57.2 2 ^{\circ} )}{31.5} = \frac{sin ( θ )}{24.4}

tan^{2} (b) sin^{2} (b) = tan^{2} (b) + sin^{2} (b)

cos (2 θ) = \frac{3}{2}, 0 \leq θ < 2 π

7.5 sin (x) = 3.4 cos (x)

What is the general solution for 1-(cos^2(x))/(1+sin(x))=2cos(x) ?
The general solution for 1-(cos^2(x))/(1+sin(x))=2cos(x) is x=1.10714…+pin