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

The general solution for (1+tan^2(x))/(1+tan^2(x))=1-sin^2(x) is x=2pin,x=pi+2pin

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

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

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

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

Switch sides

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

- sin^{2} (x) = 0

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

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

Simplify

sin^{2} (x) = 0

sin^{2} (x) = 0

Apply rule

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

sin (x) = 0

General solutions for

sin (x) = 0

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

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

θ, tan (θ) = \frac{2 3}{2}

\frac{- 3 + 4 cos ^{2} ( θ )}{1 - 2 sin ( θ )} = a + b sin (θ)

2 sin^{2} (x) - 7 sin (x) = - 3, 0 \leq x \leq 2 π

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

(218) sin^{2} (x) + (126) sin (x) - 88 = 0

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