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

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

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Popular >

1+cos(2θ)=-3cos(θ)

Solution

1 + cos (2 θ) = - 3 cos (θ)

Solution

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

Degrees

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

Solution steps

1 + cos (2 θ) = - 3 cos (θ)

Subtract

- 3 cos (θ)

from both sides

1 + cos (2 θ) + 3 cos (θ) = 0

Rewrite using trig identities

1 + cos (2 θ) + 3 cos (θ)

Use the Double Angle identity:

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

= 1 + 2 cos^{2} (θ) - 1 + 3 cos (θ)

Simplify

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

1 + 2 cos^{2} (θ) - 1 + 3 cos (θ)

Group like terms

= 2 cos^{2} (θ) + 3 cos (θ) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (θ) + 3 cos (θ)

= 2 cos^{2} (θ) + 3 cos (θ)

2 cos^{2} (θ) + 3 cos (θ) = 0

Solve by substitution

2 cos^{2} (θ) + 3 cos (θ) = 0

Let:

cos (θ) = u

2 u^{2} + 3 u = 0

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

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 \cdot 2 \cdot 0}{2 \cdot 2}

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

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

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 \cdot 2}

Separate the solutions

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

u = \frac{- 3 + 3}{2 \cdot 2} : 0

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

Add/Subtract the numbers:

- 3 + 3 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

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

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

Subtract 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}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = - \frac{3}{2}

cos (θ) = 0, cos (θ) = - \frac{3}{2}

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) = - \frac{3}{2} :

No Solution

cos (θ) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Popular Examples

tan(θ)= 6/0

tan (θ) = \frac{6}{0}

3cos(20t)+6sin(20t)=0

3 cos (20 t) + 6 sin (20 t) = 0

tan(θ)= 6/2

tan (θ) = \frac{6}{2}

tan(θ)= 6/4

tan (θ) = \frac{6}{4}

2cot^2(x)+3csc(x)=0

2 cot^{2} (x) + 3 csc (x) = 0

Frequently Asked Questions (FAQ)

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