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

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

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - cos (2 x)

from both sides

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

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

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

Simplify

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

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

- (cos (2 x) - cos^{2} (x)) : - cos (2 x) + cos^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Add similar elements:

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

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

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

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

Solve by substitution

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

Let:

cos (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 = cos (x)

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (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}

x = 0 + 2 πn

x = 0 + 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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (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}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

sin(θ)= 5/6 ,cos(θ)

sin (θ) = \frac{5}{6}, cos (θ)

tan^2(t)-sec^2(t)=1

tan^{2} (t) - sec^{2} (t) = 1

sin(x)tan(x)= 1/(cot(x))

sin (x) tan (x) = \frac{1}{cot ( x )}

0=-sqrt(2)sin(x)-1

0 = - 2 sin (x) - 1

solvefor x,z=sin(2x+3y)solve for

x, z = sin (2 x + 3 y)

Frequently Asked Questions (FAQ)

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