What is the general solution for -2cos(2θ)-5=-4sin(θ)-8 ?

The general solution for -2cos(2θ)-5=-4sin(θ)-8 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-2cos(2θ)-5=-4sin(θ)-8

Solution

- 2 cos (2 θ) - 5 = - 4 sin (θ) - 8

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

- 2 cos (2 θ) - 5 = - 4 sin (θ) - 8

Subtract

- 4 sin (θ) - 8

from both sides

4 sin (θ) - 2 cos (2 θ) + 3 = 0

Rewrite using trig identities

3 - 2 cos (2 θ) + 4 sin (θ)

Use the Double Angle identity:

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

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

Simplify

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

3 - 2 (1 - 2 sin^{2} (θ)) + 4 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)

Apply minus-plus rules

- (- a) = a

= - 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)

= 3 - 2 + 4 sin^{2} (θ) + 4 sin (θ)

Subtract the numbers:

3 - 2 = 1

= 4 sin^{2} (θ) + 4 sin (θ) + 1

= 4 sin^{2} (θ) + 4 sin (θ) + 1

1 + 4 sin (θ) + 4 sin^{2} (θ) = 0

Solve by substitution

1 + 4 sin (θ) + 4 sin^{2} (θ) = 0

Let:

sin (θ) = u

1 + 4 u + 4 u^{2} = 0

1 + 4 u + 4 u^{2} = 0 : u = - \frac{1}{2}

1 + 4 u + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} + 4 u + 1 = 0

Solve with the quadratic formula

4 u^{2} + 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 4, b = 4, c = 1

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

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

4^{2} - 4 \cdot 4 \cdot 1 = 0

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

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

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

\frac{- 4}{2 \cdot 4} = - \frac{1}{2}

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

u = - \frac{1}{2}

The solution to the quadratic equation is:

u = - \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

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 = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Popular Examples

3cos^2(θ)-cos(θ)=0

3 cos^{2} (θ) - cos (θ) = 0

10sin^2(x)+11sin(x)-8=0

10 sin^{2} (x) + 11 sin (x) - 8 = 0

2cos^2(x)+7cos(x)-4=0

2 cos^{2} (x) + 7 cos (x) - 4 = 0

cot(2θ)=(sqrt(3))/3

cot (2 θ) = \frac{3}{3}

4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

Frequently Asked Questions (FAQ)

What is the general solution for -2cos(2θ)-5=-4sin(θ)-8 ?
The general solution for -2cos(2θ)-5=-4sin(θ)-8 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin