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

The general solution for 3sin^2(θ)=2sin(θ)+3 is θ=-0.80489…+2pin,θ=pi+0.80489…+2pin

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Popular Trigonometry >

3sin^2(θ)=2sin(θ)+3

Solution

3 sin^{2} (θ) = 2 sin (θ) + 3

Solution

θ = - 0.80489 \dots + 2 πn, θ = π + 0.80489 \dots + 2 πn

Degrees

θ = - 46.11719 \dots^{\circ} + 36 0^{\circ} n, θ = 226.11719 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin^{2} (θ) = 2 sin (θ) + 3

Solve by substitution

3 sin^{2} (θ) = 2 sin (θ) + 3

Let:

sin (θ) = u

3 u^{2} = 2 u + 3

3 u^{2} = 2 u + 3 : u = \frac{1 + 10}{3}, u = \frac{1 - 10}{3}

3 u^{2} = 2 u + 3

Move

3

to the left side

3 u^{2} = 2 u + 3

Subtract

3

from both sides

3 u^{2} - 3 = 2 u + 3 - 3

Simplify

3 u^{2} - 3 = 2 u

3 u^{2} - 3 = 2 u

Move

2 u

to the left side

3 u^{2} - 3 = 2 u

Subtract

2 u

from both sides

3 u^{2} - 3 - 2 u = 2 u - 2 u

Simplify

3 u^{2} - 3 - 2 u = 0

3 u^{2} - 3 - 2 u = 0

Write in the standard form

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

3 u^{2} - 2 u - 3 = 0

Solve with the quadratic formula

3 u^{2} - 2 u - 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 2, c = - 3

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

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

(- 2)^{2} - 4 \cdot 3 (- 3) = 210

(- 2)^{2} - 4 \cdot 3 (- 3)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 3 \cdot 3

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 2^{2} + 36

2^{2} = 4

= 4 + 36

Add the numbers:

4 + 36 = 40

= 40

Prime factorization of

40 : 2^{3} \cdot 5

40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 5

Apply radical rule:

= 2^{2} 2 \cdot 5

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 5

Refine

= 210

u_{1, 2} = \frac{- ( - 2 ) \pm 2 10}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 10}{2 \cdot 3}, u_{2} = \frac{- ( - 2 ) - 2 10}{2 \cdot 3}

u = \frac{- ( - 2 ) + 2 10}{2 \cdot 3} : \frac{1 + 10}{3}

\frac{- ( - 2 ) + 2 10}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 + 2 10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

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

Factor

2 + 210 : 2 (1 + 10)

2 + 210

Rewrite as

= 2 \cdot 1 + 210

Factor out common term

2

= 2 (1 + 10)

= \frac{2 ( 1 + 10 )}{6}

Cancel the common factor:

2

= \frac{1 + 10}{3}

u = \frac{- ( - 2 ) - 2 10}{2 \cdot 3} : \frac{1 - 10}{3}

\frac{- ( - 2 ) - 2 10}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 - 2 10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

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

Factor

2 - 210 : 2 (1 - 10)

2 - 210

Rewrite as

= 2 \cdot 1 - 210

Factor out common term

2

= 2 (1 - 10)

= \frac{2 ( 1 - 10 )}{6}

Cancel the common factor:

2

= \frac{1 - 10}{3}

The solutions to the quadratic equation are:

u = \frac{1 + 10}{3}, u = \frac{1 - 10}{3}

Substitute back

u = sin (θ)

sin (θ) = \frac{1 + 10}{3}, sin (θ) = \frac{1 - 10}{3}

sin (θ) = \frac{1 + 10}{3}, sin (θ) = \frac{1 - 10}{3}

sin (θ) = \frac{1 + 10}{3} :

No Solution

sin (θ) = \frac{1 + 10}{3}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{1 - 10}{3} : θ = arcsin (\frac{1 - 10}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 10}{3}) + 2 πn

sin (θ) = \frac{1 - 10}{3}

Apply trig inverse properties

sin (θ) = \frac{1 - 10}{3}

General solutions for

sin (θ) = \frac{1 - 10}{3}

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

θ = arcsin (\frac{1 - 10}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 10}{3}) + 2 πn

θ = arcsin (\frac{1 - 10}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 10}{3}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1 - 10}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 10}{3}) + 2 πn

Show solutions in decimal form

θ = - 0.80489 \dots + 2 πn, θ = π + 0.80489 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(θ)=2sin(θ)+3 ?
The general solution for 3sin^2(θ)=2sin(θ)+3 is θ=-0.80489…+2pin,θ=pi+0.80489…+2pin