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

The general solution for (1+cos^2(x))/(sin^2(x))= 1/(sin^2(x)) is x= pi/2+2pin,x=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

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

cos^{2} (x) = 0

Apply rule

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

cos (x) = 0

General solutions for

cos (x) = 0

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

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

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

7 sin (a) = \frac{1}{11}

tan (x) = 2, x + y = 135

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

sinh (z) = - 1

1 + cos^{2} (a) = 2 cos^{2} (a)

What is the general solution for (1+cos^2(x))/(sin^2(x))= 1/(sin^2(x)) ?
The general solution for (1+cos^2(x))/(sin^2(x))= 1/(sin^2(x)) is x= pi/2+2pin,x=(3pi)/2+2pin