What is the general solution for 2cos(2θ)+7sin(θ)-2=-2 ?

The general solution for 2cos(2θ)+7sin(θ)-2=-2 is θ=-0.25268…+2pin,θ=pi+0.25268…+2pin

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2cos(2θ)+7sin(θ)-2=-2

Solution

2 cos (2 θ) + 7 sin (θ) - 2 = - 2

Solution

θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

Degrees

θ = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 194.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (2 θ) + 7 sin (θ) - 2 = - 2

Subtract

- 2

from both sides

2 cos (2 θ) + 7 sin (θ) = 0

Rewrite using trig identities

2 cos (2 θ) + 7 sin (θ)

Use the Double Angle identity:

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

= 2 (1 - 2 sin^{2} (θ)) + 7 sin (θ)

(1 - 2 sin^{2} (θ)) \cdot 2 + 7 sin (θ) = 0

Solve by substitution

(1 - 2 sin^{2} (θ)) \cdot 2 + 7 sin (θ) = 0

Let:

sin (θ) = u

(1 - 2 u^{2}) \cdot 2 + 7 u = 0

(1 - 2 u^{2}) \cdot 2 + 7 u = 0 : u = - \frac{1}{4}, u = 2

(1 - 2 u^{2}) \cdot 2 + 7 u = 0

Expand

(1 - 2 u^{2}) \cdot 2 + 7 u : 2 - 4 u^{2} + 7 u

(1 - 2 u^{2}) \cdot 2 + 7 u

= 2 (1 - 2 u^{2}) + 7 u

Expand

2 (1 - 2 u^{2}) : 2 - 4 u^{2}

2 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = 2 u^{2}

= 2 \cdot 1 - 2 \cdot 2 u^{2}

Simplify

2 \cdot 1 - 2 \cdot 2 u^{2} : 2 - 4 u^{2}

2 \cdot 1 - 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 2 = 4

= 2 - 4 u^{2}

= 2 - 4 u^{2}

= 2 - 4 u^{2} + 7 u

2 - 4 u^{2} + 7 u = 0

Write in the standard form

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

- 4 u^{2} + 7 u + 2 = 0

Solve with the quadratic formula

- 4 u^{2} + 7 u + 2 = 0

Quadratic Equation Formula:

For

a = - 4, b = 7, c = 2

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 4 ) \cdot 2}{2 ( - 4 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 4 ) \cdot 2}{2 ( - 4 )}

7^{2} - 4 (- 4) \cdot 2 = 9

7^{2} - 4 (- 4) \cdot 2

Apply rule

- (- a) = a

= 7^{2} + 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

Add the numbers:

49 + 32 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- 7 \pm 9}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 7 + 9}{2 ( - 4 )}, u_{2} = \frac{- 7 - 9}{2 ( - 4 )}

u = \frac{- 7 + 9}{2 ( - 4 )} : - \frac{1}{4}

\frac{- 7 + 9}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 + 9}{- 2 \cdot 4}

Add/Subtract the numbers:

- 7 + 9 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{- 8}

Apply the fraction rule:

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

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

u = \frac{- 7 - 9}{2 ( - 4 )} : 2

\frac{- 7 - 9}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 - 9}{- 2 \cdot 4}

Subtract the numbers:

- 7 - 9 = - 16

= \frac{- 16}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 16}{- 8}

Apply the fraction rule:

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

= \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= 2

The solutions to the quadratic equation are:

u = - \frac{1}{4}, u = 2

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{4}, sin (θ) = 2

sin (θ) = - \frac{1}{4}, sin (θ) = 2

sin (θ) = - \frac{1}{4} : θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

sin (θ) = - \frac{1}{4}

Apply trig inverse properties

sin (θ) = - \frac{1}{4}

General solutions for

sin (θ) = - \frac{1}{4}

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

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

sin (θ) = 2 :

No Solution

sin (θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ)+7sin(θ)-2=-2 ?
The general solution for 2cos(2θ)+7sin(θ)-2=-2 is θ=-0.25268…+2pin,θ=pi+0.25268…+2pin