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

The general solution for 3sin(θ)+3=2cos^2(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin

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3sin(θ)+3=2cos^2(θ)

Solution

3 sin (θ) + 3 = 2 cos^{2} (θ)

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (θ) + 3 = 2 cos^{2} (θ)

Subtract

2 cos^{2} (θ)

from both sides

3 sin (θ) + 3 - 2 cos^{2} (θ) = 0

Rewrite using trig identities

3 - 2 cos^{2} (θ) + 3 sin (θ)

Use the Pythagorean identity:

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

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

= 3 - 2 (1 - sin^{2} (θ)) + 3 sin (θ)

Simplify

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

3 - 2 (1 - sin^{2} (θ)) + 3 sin (θ)

Expand

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

- 2 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (θ)

= - 2 \cdot 1 - (- 2) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 sin^{2} (θ)

= 3 - 2 + 2 sin^{2} (θ) + 3 sin (θ)

Subtract the numbers:

3 - 2 = 1

= 2 sin^{2} (θ) + 3 sin (θ) + 1

= 2 sin^{2} (θ) + 3 sin (θ) + 1

1 + 2 sin^{2} (θ) + 3 sin (θ) = 0

Solve by substitution

1 + 2 sin^{2} (θ) + 3 sin (θ) = 0

Let:

sin (θ) = u

1 + 2 u^{2} + 3 u = 0

1 + 2 u^{2} + 3 u = 0 : u = - \frac{1}{2}, u = - 1

1 + 2 u^{2} + 3 u = 0

Write in the standard form

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

2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

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

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

3^{2} - 4 \cdot 2 \cdot 1 = 1

3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 \cdot 2}, u_{2} = \frac{- 3 - 1}{2 \cdot 2}

u = \frac{- 3 + 1}{2 \cdot 2} : - \frac{1}{2}

\frac{- 3 + 1}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 3 - 1}{2 \cdot 2} : - 1

\frac{- 3 - 1}{2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{2}, sin (θ) = - 1

sin (θ) = - \frac{1}{2}, sin (θ) = - 1

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

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Popular Examples

cos(x)=-5/6

cos (x) = - \frac{5}{6}

2sin(x/2)-sqrt(3)=0

2 sin (\frac{x}{2}) - 3 = 0

2sin^2(θ)=sin(θ)+1

2 sin^{2} (θ) = sin (θ) + 1

2sec^2(x)-4=0

2 sec^{2} (x) - 4 = 0

tan(a)=sqrt(3)

tan (a) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 3sin(θ)+3=2cos^2(θ) ?
The general solution for 3sin(θ)+3=2cos^2(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin