What is the general solution for sqrt(3)=sqrt(3)cos(2θ)+sin(2θ) ?

The general solution for sqrt(3)=sqrt(3)cos(2θ)+sin(2θ) is θ=2pin,θ=pi+2pin,θ= pi/6+pin

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sqrt(3)=sqrt(3)cos(2θ)+sin(2θ)

Solution

3 = 3 cos (2 θ) + sin (2 θ)

Solution

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

3 = 3 cos (2 θ) + sin (2 θ)

Switch sides

3 cos (2 θ) + sin (2 θ) = 3

Subtract

3

from both sides

3 cos (2 θ) + sin (2 θ) - 3 = 0

Rewrite using trig identities

sin (2 θ) - 3 + cos (2 θ) 3

Use the Double Angle identity:

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

= 2 sin (θ) cos (θ) - 3 + 3 cos (2 θ)

Use the Double Angle identity:

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

= - 3 + 3 (1 - 2 sin^{2} (θ)) + 2 cos (θ) sin (θ)

Simplify

- 3 + 3 (1 - 2 sin^{2} (θ)) + 2 cos (θ) sin (θ) : - 23 sin^{2} (θ) + 2 cos (θ) sin (θ)

- 3 + 3 (1 - 2 sin^{2} (θ)) + 2 cos (θ) sin (θ)

Expand

3 (1 - 2 sin^{2} (θ)) : 3 - 23 sin^{2} (θ)

3 (1 - 2 sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = 2 sin^{2} (θ)

= 3 \cdot 1 - 3 \cdot 2 sin^{2} (θ)

= 1 \cdot 3 - 23 sin^{2} (θ)

Multiply:

1 \cdot 3 = 3

= 3 - 23 sin^{2} (θ)

= - 3 + 3 - 23 sin^{2} (θ) + 2 cos (θ) sin (θ)

Add similar elements:

- 3 + 3 = 0

= - 23 sin^{2} (θ) + 2 cos (θ) sin (θ)

= - 23 sin^{2} (θ) + 2 cos (θ) sin (θ)

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

Factor

2 cos (θ) sin (θ) - 2 sin^{2} (θ) 3 : 2 sin (θ) (cos (θ) - 3 sin (θ))

2 cos (θ) sin (θ) - 2 sin^{2} (θ) 3

Apply exponent rule:

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

sin^{2} (θ) = sin (θ) sin (θ)

= 2 cos (θ) sin (θ) - 2 sin (θ) sin (θ) 3

Rewrite as

= 2 sin (θ) cos (θ) - 2 sin (θ) sin (θ) 3

Factor out common term

2 sin (θ)

= 2 sin (θ) (cos (θ) - sin (θ) 3)

2 sin (θ) (cos (θ) - 3 sin (θ)) = 0

Solving each part separately

sin (θ) = 0 or cos (θ) - 3 sin (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 (θ) - 3 sin (θ) = 0 : θ = \frac{π}{6} + πn

cos (θ) - 3 sin (θ) = 0

Rewrite using trig identities

cos (θ) - 3 sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{cos ( θ ) - 3 sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

1 - \frac{3 sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

1 - 3 tan (θ) = 0

1 - 3 tan (θ) = 0

Move

1

to the right side

1 - 3 tan (θ) = 0

Subtract

1

from both sides

1 - 3 tan (θ) - 1 = 0 - 1

Simplify

- 3 tan (θ) = - 1

- 3 tan (θ) = - 1

Divide both sides by

- 3

- 3 tan (θ) = - 1

Divide both sides by

- 3

\frac{- 3 tan ( θ )}{- 3} = \frac{- 1}{- 3}

Simplify

\frac{- 3 tan ( θ )}{- 3} = \frac{- 1}{- 3}

Simplify

\frac{- 3 tan ( θ )}{- 3} : tan (θ)

\frac{- 3 tan ( θ )}{- 3}

Apply the fraction rule:

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

= \frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

\frac{- 1}{- 3} : \frac{3}{3}

\frac{- 1}{- 3}

Apply the fraction rule:

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

= \frac{1}{3}

Rationalize

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

= \frac{3}{3}

tan (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \frac{3}{3}

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}{6} + πn

θ = \frac{π}{6} + πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + πn

Popular Examples

8sin^2(x)=sin(x)

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

6cos(x)+6sin(x)=3sqrt(6)

6 cos (x) + 6 sin (x) = 36

3sin(x)+1=0

3 sin (x) + 1 = 0

(tan(x)+1)(cos(x)-1)=0

(tan (x) + 1) (cos (x) - 1) = 0

cot(θ)=-1/(sqrt(3))

cot (θ) = - \frac{1}{3}

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)=sqrt(3)cos(2θ)+sin(2θ) ?
The general solution for sqrt(3)=sqrt(3)cos(2θ)+sin(2θ) is θ=2pin,θ=pi+2pin,θ= pi/6+pin