What is the general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) ?

The general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) is b=2pin,b=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

b = 2 πn, b = π + 2 πn

Solution steps

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

Subtract

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

from both sides

tan^{2} (b) sin^{2} (b) - tan^{2} (b) - sin^{2} (b) = 0

Express with sin, cos

\frac{- sin ^{2} ( b ) + sin ^{4} ( b ) - cos ^{2} ( b ) sin ^{2} ( b )}{cos ^{2} ( b )} = 0

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

- sin^{2} (b) + sin^{4} (b) - cos^{2} (b) sin^{2} (b) = 0

Factor

- sin^{2} (b) + sin^{4} (b) - cos^{2} (b) sin^{2} (b) : - sin^{2} (b) (1 - sin^{2} (b) + cos^{2} (b))

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

Solving each part separately

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

sin^{2} (b) = 0 : b = 2 πn, b = π + 2 πn

1 - sin^{2} (b) + cos^{2} (b) = 0 : b = \frac{π}{2} + πn

Combine all the solutions

b = 2 πn, b = π + 2 πn, b = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

b = 2 πn, b = π + 2 πn

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

7.5 sin (x) = 3.4 cos (x)

sec (θ) = 8, tan (θ) = 37

tan (3 x - 4) = \frac{- 1}{2}

π cos (2 θ) = 0

What is the general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) ?
The general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) is b=2pin,b=pi+2pin