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

The general solution for (cos^2(x)-1)/(sin^2(x))=-tan(x) is x= pi/4+pin

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

Solution

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

Solution

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

- tan (x)

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Factor

- sin^{2} (x) + sin^{2} (x) tan (x) : sin^{2} (x) (tan (x) - 1)

- sin^{2} (x) + sin^{2} (x) tan (x)

Factor out common term

sin^{2} (x)

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

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

Solving each part separately

sin^{2} (x) = 0 or tan (x) - 1 = 0

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

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

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

tan (x) - 1 + 1 = 0 + 1

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

Since the equation is undefined for:

2 πn, π + 2 πn

x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for (cos^2(x)-1)/(sin^2(x))=-tan(x) ?
The general solution for (cos^2(x)-1)/(sin^2(x))=-tan(x) is x= pi/4+pin