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

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

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

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

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

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

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

Rewrite using trig identities

1 + cos^{2} (x) - sin^{2} (x)

Use the Double Angle identity:

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

= 1 + cos (2 x)

1 + cos (2 x) = 0

Move

1

to the right side

1 + cos (2 x) = 0

Subtract

1

from both sides

1 + cos (2 x) - 1 = 0 - 1

Simplify

cos (2 x) = - 1

cos (2 x) = - 1

General solutions for

cos (2 x) = - 1

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}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

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

w, 100 cos (10 w) = 50

cos (θ) = \frac{8}{25}

sin (2 x) = π

sin (x + 5 π) = 0

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