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

The general solution for sin(x)=1-cos^x(x) is x=2pin

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sin(x)=1-cos^x(x)

Solution

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

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Square both sides

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

Subtract

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

from both sides

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

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

plug in

x = 2 π 1

sin (2 π 1) = 1 - cos^{2 π 1} (2 π 1)

Refine

0 = 0

\Rightarrow True

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

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

plug in

x = π + 2 π 1

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

Undefined

\Rightarrow False

x = 2 πn

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Frequently Asked Questions (FAQ)

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