English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

2sin(2θ)cos(θ)+cos(2θ)sin(θ)=0

Solution

2 sin (2 θ) cos (θ) + cos (2 θ) sin (θ) = 0

Solution

θ = 2 πn, θ = π + 2 πn, θ = - 1.15026 \dots + πn, θ = 1.15026 \dots + πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = - 65.90515 \dots^{\circ} + 18 0^{\circ} n, θ = 65.90515 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 sin (2 θ) cos (θ) + cos (2 θ) sin (θ) = 0

Rewrite using trig identities

cos (2 θ) sin (θ) + 2 cos (θ) sin (2 θ)

Use the Double Angle identity:

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

= cos (2 θ) sin (θ) + 2 cos (θ) \cdot 2 sin (θ) cos (θ)

2 cos (θ) \cdot 2 sin (θ) cos (θ) = 4 cos^{2} (θ) sin (θ)

2 cos (θ) \cdot 2 sin (θ) cos (θ)

Multiply the numbers:

2 \cdot 2 = 4

= 4 cos (θ) sin (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 4 sin (θ) cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 4 sin (θ) cos^{2} (θ)

= cos (2 θ) sin (θ) + 4 cos^{2} (θ) sin (θ)

cos (2 θ) sin (θ) + 4 cos^{2} (θ) sin (θ) = 0

Factor

cos (2 θ) sin (θ) + 4 cos^{2} (θ) sin (θ) : sin (θ) (cos (2 θ) + 4 cos^{2} (θ))

cos (2 θ) sin (θ) + 4 cos^{2} (θ) sin (θ)

Factor out common term

sin (θ)

= sin (θ) (cos (2 θ) + 4 cos^{2} (θ))

sin (θ) (cos (2 θ) + 4 cos^{2} (θ)) = 0

Solving each part separately

sin (θ) = 0 or cos (2 θ) + 4 cos^{2} (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (2 θ) + 4 cos^{2} (θ) = 0 : θ = arctan (- 5) + πn, θ = arctan (5) + πn

cos (2 θ) + 4 cos^{2} (θ) = 0

Rewrite using trig identities

cos (2 θ) + 4 cos^{2} (θ)

Use the Double Angle identity:

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

= cos^{2} (θ) - sin^{2} (θ) + 4 cos^{2} (θ)

Simplify

= 5 cos^{2} (θ) - sin^{2} (θ)

- sin^{2} (θ) + 5 cos^{2} (θ) = 0

Factor

- sin^{2} (θ) + 5 cos^{2} (θ) : (5 cos (θ) + sin (θ)) (5 cos (θ) - sin (θ))

- sin^{2} (θ) + 5 cos^{2} (θ)

Rewrite

5 cos^{2} (θ) - sin^{2} (θ)

(5 cos (θ))^{2} - sin^{2} (θ)

5 cos^{2} (θ) - sin^{2} (θ)

Apply radical rule:

a = (a)^{2}

5 = (5)^{2}

= (5)^{2} cos^{2} (θ) - sin^{2} (θ)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(5)^{2} cos^{2} (θ) = (5 cos (θ))^{2}

= (5 cos (θ))^{2} - sin^{2} (θ)

= (5 cos (θ))^{2} - sin^{2} (θ)

Apply Difference of Two Squares Formula:

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

(5 cos (θ))^{2} - sin^{2} (θ) = (5 cos (θ) + sin (θ)) (5 cos (θ) - sin (θ))

= (5 cos (θ) + sin (θ)) (5 cos (θ) - sin (θ))

(5 cos (θ) + sin (θ)) (5 cos (θ) - sin (θ)) = 0

Solving each part separately

5 cos (θ) + sin (θ) = 0 or 5 cos (θ) - sin (θ) = 0

5 cos (θ) + sin (θ) = 0 : θ = arctan (- 5) + πn

5 cos (θ) + sin (θ) = 0

Rewrite using trig identities

5 cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{5 cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

5 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

5 + tan (θ) = 0

5 + tan (θ) = 0

Move

5

to the right side

5 + tan (θ) = 0

Subtract

5

from both sides

5 + tan (θ) - 5 = 0 - 5

Simplify

tan (θ) = - 5

tan (θ) = - 5

Apply trig inverse properties

tan (θ) = - 5

General solutions for

tan (θ) = - 5

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- 5) + πn

θ = arctan (- 5) + πn

5 cos (θ) - sin (θ) = 0 : θ = arctan (5) + πn

5 cos (θ) - sin (θ) = 0

Rewrite using trig identities

5 cos (θ) - sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{5 cos ( θ ) - sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

5 - \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

5 - tan (θ) = 0

5 - tan (θ) = 0

Move

5

to the right side

5 - tan (θ) = 0

Subtract

5

from both sides

5 - tan (θ) - 5 = 0 - 5

Simplify

- tan (θ) = - 5

- tan (θ) = - 5

Divide both sides by

- 1

- tan (θ) = - 5

Divide both sides by

- 1

\frac{- tan ( θ )}{- 1} = \frac{- 5}{- 1}

Simplify

tan (θ) = 5

tan (θ) = 5

Apply trig inverse properties

tan (θ) = 5

General solutions for

tan (θ) = 5

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (5) + πn

θ = arctan (5) + πn

Combine all the solutions

θ = arctan (- 5) + πn, θ = arctan (5) + πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arctan (- 5) + πn, θ = arctan (5) + πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = - 1.15026 \dots + πn, θ = 1.15026 \dots + πn

Graph

Popular Examples

2sin^2(x)=2+cos(x),0<= x<= 2pi