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

The general solution for tan^2(x)-cos^2(x)=sin^2(x) is x= pi/4+pin,x=(3pi)/4+pin

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Popular >

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

Solution

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

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= tan^{2} (x) - 1

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

Solve by substitution

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

Let:

tan (x) = u

- 1 + u^{2} = 0

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

- 1 + u^{2} + 1 = 0 + 1

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - 1

tan (x) = 1, tan (x) = - 1

tan (x) = 1 : x = \frac{π}{4} + πn

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

tan (x) = - 1 : x = \frac{3 π}{4} + πn

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

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cos^{2} (x) = \frac{( 8 k - 7 )}{7}

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2 cos^{2} (x) - 7 cos (x) = - 3

sin(x)+2sin(2x)=0

sin (x) + 2 sin (2 x) = 0

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

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cosh (x) cosh (1) - sinh (x) sinh (1) = 2

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-cos^2(x)=sin^2(x) ?
The general solution for tan^2(x)-cos^2(x)=sin^2(x) is x= pi/4+pin,x=(3pi)/4+pin