What is the general solution for sin(2θ)=2cos^2(θ) ?

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

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sin(2θ)=2cos^2(θ)

Solution

sin (2 θ) = 2 cos^{2} (θ)

Solution

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{4} + πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 θ) = 2 cos^{2} (θ)

Subtract

2 cos^{2} (θ)

from both sides

sin (2 θ) - 2 cos^{2} (θ) = 0

Rewrite using trig identities

sin (2 θ) - 2 cos^{2} (θ)

Use the Double Angle identity:

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

= 2 sin (θ) cos (θ) - 2 cos^{2} (θ)

- 2 cos^{2} (θ) + 2 cos (θ) sin (θ) = 0

Factor

- 2 cos^{2} (θ) + 2 cos (θ) sin (θ) : 2 cos (θ) (- cos (θ) + sin (θ))

- 2 cos^{2} (θ) + 2 cos (θ) sin (θ)

Apply exponent rule:

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

cos^{2} (θ) = cos (θ) cos (θ)

= - 2 cos (θ) cos (θ) + 2 sin (θ) cos (θ)

Factor out common term

2 cos (θ)

= 2 cos (θ) (- cos (θ) + sin (θ))

2 cos (θ) (- cos (θ) + sin (θ)) = 0

Solving each part separately

cos (θ) = 0 or - cos (θ) + sin (θ) = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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

- cos (θ) + sin (θ) = 0

Rewrite using trig identities

- cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{- cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

- 1 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

- 1 + tan (θ) = 0

- 1 + tan (θ) = 0

Move

1

to the right side

- 1 + tan (θ) = 0

Add

1

to both sides

- 1 + tan (θ) + 1 = 0 + 1

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{4} + πn

Popular Examples

0=3cos(3θ)

0 = 3 cos (3 θ)

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

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

tan(2x-pi/4)=-1

tan (2 x - \frac{π}{4}) = - 1

2sin^2(A)=5sin(A)-2

2 sin^{2} (A) = 5 sin (A) - 2

cot(θ)= 12/5

cot (θ) = \frac{12}{5}

Frequently Asked Questions (FAQ)

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