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

The general solution for 2cos(2θ)-2=-3sin(θ) is θ=2pin,θ=pi+2pin,θ=0.84806…+2pin,θ=pi-0.84806…+2pin

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2cos(2θ)-2=-3sin(θ)

Solution

2 cos (2 θ) - 2 = - 3 sin (θ)

Solution

θ = 2 πn, θ = π + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 48.59037 \dots^{\circ} + 36 0^{\circ} n, θ = 131.40962 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (2 θ) - 2 = - 3 sin (θ)

Subtract

- 3 sin (θ)

from both sides

2 cos (2 θ) - 2 + 3 sin (θ) = 0

Rewrite using trig identities

- 2 + 2 cos (2 θ) + 3 sin (θ)

Use the Double Angle identity:

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

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

Simplify

- 2 + 2 (1 - 2 sin^{2} (θ)) + 3 sin (θ) : 3 sin (θ) - 4 sin^{2} (θ)

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

Expand

2 (1 - 2 sin^{2} (θ)) : 2 - 4 sin^{2} (θ)

2 (1 - 2 sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = 2 sin^{2} (θ)

= 2 \cdot 1 - 2 \cdot 2 sin^{2} (θ)

Simplify

2 \cdot 1 - 2 \cdot 2 sin^{2} (θ) : 2 - 4 sin^{2} (θ)

2 \cdot 1 - 2 \cdot 2 sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (θ)

Multiply the numbers:

2 \cdot 2 = 4

= 2 - 4 sin^{2} (θ)

= 2 - 4 sin^{2} (θ)

= - 2 + 2 - 4 sin^{2} (θ) + 3 sin (θ)

- 2 + 2 = 0

= 3 sin (θ) - 4 sin^{2} (θ)

= 3 sin (θ) - 4 sin^{2} (θ)

3 sin (θ) - 4 sin^{2} (θ) = 0

Solve by substitution

3 sin (θ) - 4 sin^{2} (θ) = 0

Let:

sin (θ) = u

3 u - 4 u^{2} = 0

3 u - 4 u^{2} = 0 : u = 0, u = \frac{3}{4}

3 u - 4 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = 3, c = 0

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

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

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

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

Apply rule

- (- a) = a

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

Separate the solutions

u_{1} = \frac{- 3 + 3}{2 ( - 4 )}, u_{2} = \frac{- 3 - 3}{2 ( - 4 )}

u = \frac{- 3 + 3}{2 ( - 4 )} : 0

\frac{- 3 + 3}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 3}{- 2 \cdot 4}

Add/Subtract the numbers:

- 3 + 3 = 0

= \frac{0}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{0}{- 8}

Apply the fraction rule:

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

= - \frac{0}{8}

Apply rule

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

= - 0

= 0

u = \frac{- 3 - 3}{2 ( - 4 )} : \frac{3}{4}

\frac{- 3 - 3}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 3}{- 2 \cdot 4}

Subtract the numbers:

- 3 - 3 = - 6

= \frac{- 6}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 6}{- 8}

Apply the fraction rule:

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

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

The solutions to the quadratic equation are:

u = 0, u = \frac{3}{4}

Substitute back

u = sin (θ)

sin (θ) = 0, sin (θ) = \frac{3}{4}

sin (θ) = 0, sin (θ) = \frac{3}{4}

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

sin (θ) = \frac{3}{4} : θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

sin (θ) = \frac{3}{4}

Apply trig inverse properties

sin (θ) = \frac{3}{4}

General solutions for

sin (θ) = \frac{3}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

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csc (θ) = \frac{7}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ)-2=-3sin(θ) ?
The general solution for 2cos(2θ)-2=-3sin(θ) is θ=2pin,θ=pi+2pin,θ=0.84806…+2pin,θ=pi-0.84806…+2pin