What is the general solution for cos(2θ)=1+3sin(θ) ?

The general solution for cos(2θ)=1+3sin(θ) is θ=2pin,θ=pi+2pin

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cos(2θ)=1+3sin(θ)

Solution

cos (2 θ) = 1 + 3 sin (θ)

Solution

θ = 2 πn, θ = π + 2 πn

Degrees

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

Solution steps

cos (2 θ) = 1 + 3 sin (θ)

Subtract

1 + 3 sin (θ)

from both sides

cos (2 θ) - 1 - 3 sin (θ) = 0

Rewrite using trig identities

- 1 + cos (2 θ) - 3 sin (θ)

Use the Double Angle identity:

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

= - 1 + 1 - 2 sin^{2} (θ) - 3 sin (θ)

Simplify

= - 2 sin^{2} (θ) - 3 sin (θ)

- 2 sin^{2} (θ) - 3 sin (θ) = 0

Solve by substitution

- 2 sin^{2} (θ) - 3 sin (θ) = 0

Let:

sin (θ) = u

- 2 u^{2} - 3 u = 0

- 2 u^{2} - 3 u = 0 : u = - \frac{3}{2}, u = 0

- 2 u^{2} - 3 u = 0

Solve with the quadratic formula

- 2 u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = - 2, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 0 = 3

(- 3)^{2} - 4 (- 2) \cdot 0

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 2 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} + 0

3^{2} + 0 = 3^{2}

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 3}{2 ( - 2 )} : - \frac{3}{2}

\frac{- ( - 3 ) + 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{3 + 3}{- 2 \cdot 2}

Add the numbers:

3 + 3 = 6

= \frac{6}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{- 4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

u = \frac{- ( - 3 ) - 3}{2 ( - 2 )} : 0

\frac{- ( - 3 ) - 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{3 - 3}{- 2 \cdot 2}

Subtract the numbers:

3 - 3 = 0

= \frac{0}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

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

= - \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{3}{2}, u = 0

Substitute back

u = sin (θ)

sin (θ) = - \frac{3}{2}, sin (θ) = 0

sin (θ) = - \frac{3}{2}, sin (θ) = 0

sin (θ) = - \frac{3}{2} :

No Solution

sin (θ) = - \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

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

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

Popular Examples

cos(2x)-cos(x)=-1,0<= x<2pi

cos (2 x) - cos (x) = - 1, 0 \leq x < 2 π

sin(x)=-1/(sqrt(10))

sin (x) = - \frac{1}{10}

sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π

6sec^2(x)=0

6 sec^{2} (x) = 0

sqrt(3-2cos(x))=2

3 - 2 cos (x) = 2

Frequently Asked Questions (FAQ)

What is the general solution for cos(2θ)=1+3sin(θ) ?
The general solution for cos(2θ)=1+3sin(θ) is θ=2pin,θ=pi+2pin