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

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

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cos(x)=(2tan(x))/(1+tan^2(x))

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

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

Solution steps

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

Subtract

\frac{2 tan ( x )}{1 + tan ^{2} ( x )}

from both sides

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

Simplify

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

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

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

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

Express with sin, cos

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

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

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

Rewrite using trig identities

- 2 sin (x) + 1 = 0

Move

1

to the right side

- 2 sin (x) = - 1

Divide both sides by

- 2

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

General solutions for

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

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

(1 + cos (4 x)) sin (2 x) = cos^{2} (2 x)

cos (2 x) - cos (6 x) - sin (4 x) = 0

cot (θ) = \frac{15}{8}

2 sin (2 x) = sin (4 x)

u, sin (u) = x

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