What is the general solution for -3cos(2θ)+3=5sin(θ) ?

The general solution for -3cos(2θ)+3=5sin(θ) is θ=0.98511…+2pin,θ=pi-0.98511…+2pin,θ=2pin,θ=pi+2pin

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Popular >

-3cos(2θ)+3=5sin(θ)

Solution

- 3 cos (2 θ) + 3 = 5 sin (θ)

Solution

θ = 0.98511 \dots + 2 πn, θ = π - 0.98511 \dots + 2 πn, θ = 2 πn, θ = π + 2 πn

Degrees

θ = 56.44269 \dots^{\circ} + 36 0^{\circ} n, θ = 123.55730 \dots^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- 3 cos (2 θ) + 3 = 5 sin (θ)

Subtract

5 sin (θ)

from both sides

- 3 cos (2 θ) + 3 - 5 sin (θ) = 0

Rewrite using trig identities

3 - 3 cos (2 θ) - 5 sin (θ)

Use the Double Angle identity:

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

= 3 - 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ)

Simplify

3 - 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ) : 6 sin^{2} (θ) - 5 sin (θ)

3 - 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ)

Expand

- 3 (1 - 2 sin^{2} (θ)) : - 3 + 6 sin^{2} (θ)

- 3 (1 - 2 sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = 2 sin^{2} (θ)

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

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

Simplify

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ) : - 3 + 6 sin^{2} (θ)

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 sin^{2} (θ)

Multiply the numbers:

3 \cdot 2 = 6

= - 3 + 6 sin^{2} (θ)

= - 3 + 6 sin^{2} (θ)

= 3 - 3 + 6 sin^{2} (θ) - 5 sin (θ)

3 - 3 = 0

= 6 sin^{2} (θ) - 5 sin (θ)

= 6 sin^{2} (θ) - 5 sin (θ)

- 5 sin (θ) + 6 sin^{2} (θ) = 0

Solve by substitution

- 5 sin (θ) + 6 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 5 u + 6 u^{2} = 0

- 5 u + 6 u^{2} = 0 : u = \frac{5}{6}, u = 0

- 5 u + 6 u^{2} = 0

Write in the standard form

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

6 u^{2} - 5 u = 0

Solve with the quadratic formula

6 u^{2} - 5 u = 0

Quadratic Equation Formula:

For

a = 6, b = - 5, c = 0

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 \cdot 0 = 5

(- 5)^{2} - 4 \cdot 6 \cdot 0

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 6 \cdot 0

Apply rule

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

Apply radical rule:

assuming

a \geq 0

= 5

u_{1, 2} = \frac{- ( - 5 ) \pm 5}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 5}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 5}{2 \cdot 6}

u = \frac{- ( - 5 ) + 5}{2 \cdot 6} : \frac{5}{6}

\frac{- ( - 5 ) + 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 + 5}{2 \cdot 6}

Add the numbers:

5 + 5 = 10

= \frac{10}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{10}{12}

Cancel the common factor:

2

= \frac{5}{6}

u = \frac{- ( - 5 ) - 5}{2 \cdot 6} : 0

\frac{- ( - 5 ) - 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 - 5}{2 \cdot 6}

Subtract the numbers:

5 - 5 = 0

= \frac{0}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{0}{12}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{5}{6}, u = 0

Substitute back

u = sin (θ)

sin (θ) = \frac{5}{6}, sin (θ) = 0

sin (θ) = \frac{5}{6}, sin (θ) = 0

sin (θ) = \frac{5}{6} : θ = arcsin (\frac{5}{6}) + 2 πn, θ = π - arcsin (\frac{5}{6}) + 2 πn

sin (θ) = \frac{5}{6}

Apply trig inverse properties

sin (θ) = \frac{5}{6}

General solutions for

sin (θ) = \frac{5}{6}

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

θ = arcsin (\frac{5}{6}) + 2 πn, θ = π - arcsin (\frac{5}{6}) + 2 πn

θ = arcsin (\frac{5}{6}) + 2 πn, θ = π - arcsin (\frac{5}{6}) + 2 πn

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

Combine all the solutions

θ = arcsin (\frac{5}{6}) + 2 πn, θ = π - arcsin (\frac{5}{6}) + 2 πn, θ = 2 πn, θ = π + 2 πn

Show solutions in decimal form

θ = 0.98511 \dots + 2 πn, θ = π - 0.98511 \dots + 2 πn, θ = 2 πn, θ = π + 2 πn

Popular Examples

12sin^2(x)+18sin(x)+6=0

12 sin^{2} (x) + 18 sin (x) + 6 = 0

csc^2(θ)+cot(θ)=0

csc^{2} (θ) + cot (θ) = 0

sin^2(θ)-2sin(θ)-3=0

sin^{2} (θ) - 2 sin (θ) - 3 = 0

5tan(θ)-4=0

5 tan (θ) - 4 = 0

0=cos(x/2)

0 = cos (\frac{x}{2})

Frequently Asked Questions (FAQ)

What is the general solution for -3cos(2θ)+3=5sin(θ) ?
The general solution for -3cos(2θ)+3=5sin(θ) is θ=0.98511…+2pin,θ=pi-0.98511…+2pin,θ=2pin,θ=pi+2pin