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

The general solution for (1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1) is θ=0.47366…+pin,θ=-0.47366…+pin

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(1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1)

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

θ = 0.47366 \dots + πn, θ = - 0.47366 \dots + πn

Solution steps

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

cos (- 1) = cos (1)

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

Solve by substitution

tan (θ) = \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}, tan (θ) = - \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}

tan (θ) = \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1} : θ = arctan (\frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

tan (θ) = - \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1} : θ = arctan (- \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

Combine all the solutions

θ = arctan (\frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn, θ = arctan (- \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

Show solutions in decimal form

θ = 0.47366 \dots + πn, θ = - 0.47366 \dots + πn

sin (x) = - \frac{5}{6}

5 cot (x) + 3 = 8

cos (x) = - \frac{3}{2}, 0 \leq x \leq 2 π

9 sin (θ) + 2 = 0

2 sin (t) = 0

What is the general solution for (1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1) ?
The general solution for (1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1) is θ=0.47366…+pin,θ=-0.47366…+pin