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

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

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Popular >

cos(θ)sin(θ)=sin(2θ)

Solution

cos (θ) sin (θ) = sin (2 θ)

Solution

θ = πn, θ = \frac{π}{2} + πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos (θ) sin (θ) = sin (2 θ)

Subtract

sin (2 θ)

from both sides

cos (θ) sin (θ) - sin (2 θ) = 0

Rewrite using trig identities

- sin (2 θ) + cos (θ) sin (θ)

Use the Double Angle identity:

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

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

= - sin (2 θ) + \frac{sin ( 2 θ )}{2}

Simplify

- sin (2 θ) + \frac{sin ( 2 θ )}{2} : - \frac{sin ( 2 θ )}{2}

- sin (2 θ) + \frac{sin ( 2 θ )}{2}

Convert element to fraction:

sin (2 θ) = \frac{sin ( 2 θ ) 2}{2}

= \frac{sin ( 2 θ )}{2} - \frac{sin ( 2 θ ) \cdot 2}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( 2 θ ) - sin ( 2 θ ) \cdot 2}{2}

Add similar elements:

sin (2 θ) - 2 sin (2 θ) = - sin (2 θ)

= \frac{- sin ( 2 θ )}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{sin ( 2 θ )}{2}

= - \frac{sin ( 2 θ )}{2}

- \frac{sin ( 2 θ )}{2} = 0

Multiply both sides by

2

- \frac{sin ( 2 θ )}{2} = 0

Multiply both sides by

2

2 (- \frac{sin ( 2 θ )}{2}) = 0 \cdot 2

Simplify

- sin (2 θ) = 0

- sin (2 θ) = 0

Divide both sides by

- 1

- sin (2 θ) = 0

Divide both sides by

- 1

\frac{- sin ( 2 θ )}{- 1} = \frac{0}{- 1}

Simplify

sin (2 θ) = 0

sin (2 θ) = 0

General solutions for

sin (2 θ) = 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}

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

Divide both sides by

2

2 θ = 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2}

Simplify

θ = πn

θ = πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

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

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = πn, θ = \frac{π}{2} + πn

Popular Examples

4sin(x)cos(x)=-3cos(x)

4 sin (x) cos (x) = - 3 cos (x)

3sin^2(x)-cos^2(x)=0

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

solvefor tan(θ)=(1.0667)/(0.5),θ solve for

tan (θ) = \frac{1.0667}{0.5}, θ

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

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

2=sec(x)

2 = sec (x)

Frequently Asked Questions (FAQ)

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