What is the general solution for -2cos(2θ)+12=4sin(θ)+9 ?

The general solution for -2cos(2θ)+12=4sin(θ)+9 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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Popular >

-2cos(2θ)+12=4sin(θ)+9

Solution

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

Subtract

4 sin (θ) + 9

from both sides

- 2 cos (2 θ) - 4 sin (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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

= 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}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \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{π}{6} + 2 πn, θ = \frac{5 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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Popular Examples

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(cot(x)-1)(2sin(x)+1)=0

(cot (x) - 1) (2 sin (x) + 1) = 0

solvefor y,x=tan(y)solve for

y, x = tan (y)

tan(x)=-12/5

tan (x) = - \frac{12}{5}

2cos(x)=sin(x)

2 cos (x) = sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for -2cos(2θ)+12=4sin(θ)+9 ?
The general solution for -2cos(2θ)+12=4sin(θ)+9 is θ= pi/6+2pin,θ=(5pi)/6+2pin