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

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

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

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sin (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

1 - 1 = 0

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

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

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

Factor

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

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

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

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

Factor out common term

2 sin (x)

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

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

Solving each part separately

sin (x) = 0 or sin (x) - cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

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

sin (x) - cos (x) = 0 : x = \frac{π}{4} + πn

sin (x) - cos (x) = 0

Rewrite using trig identities

sin (x) - cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{sin ( x ) - cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) - 1 = 0

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

Popular Examples

cos(θ)=5

cos (θ) = 5

2cos^2(x)+2cos(x)=0

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

solvefor x,tan(3x)=5tan(x)solve for

x, tan (3 x) = 5 tan (x)

3sin(2t)=0

3 sin (2 t) = 0

tan(θ)= 8/5

tan (θ) = \frac{8}{5}

Frequently Asked Questions (FAQ)

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