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

The general solution for 3sin^2(θ)-2sin(θ)=2 is θ=-0.58066…+2pin,θ=pi+0.58066…+2pin

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Popular >

3sin^2(θ)-2sin(θ)=2

Solution

3 sin^{2} (θ) - 2 sin (θ) = 2

Solution

θ = - 0.58066 \dots + 2 πn, θ = π + 0.58066 \dots + 2 πn

Degrees

θ = - 33.26990 \dots^{\circ} + 36 0^{\circ} n, θ = 213.26990 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin^{2} (θ) - 2 sin (θ) = 2

Solve by substitution

3 sin^{2} (θ) - 2 sin (θ) = 2

Let:

sin (θ) = u

3 u^{2} - 2 u = 2

3 u^{2} - 2 u = 2 : u = \frac{1 + 7}{3}, u = \frac{1 - 7}{3}

3 u^{2} - 2 u = 2

Move

2

to the left side

3 u^{2} - 2 u = 2

Subtract

2

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 \cdot 3 (- 2) = 27

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 2^{2} + 24

2^{2} = 4

= 4 + 24

Add the numbers:

4 + 24 = 28

= 28

Prime factorization of

28 : 2^{2} \cdot 7

28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

Apply radical rule:

= 7 2^{2}

Apply radical rule:

2^{2} = 2

= 27

u_{1, 2} = \frac{- ( - 2 ) \pm 2 7}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 7}{2 \cdot 3}, u_{2} = \frac{- ( - 2 ) - 2 7}{2 \cdot 3}

u = \frac{- ( - 2 ) + 2 7}{2 \cdot 3} : \frac{1 + 7}{3}

\frac{- ( - 2 ) + 2 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 + 2 7}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2 + 2 7}{6}

Factor

2 + 27 : 2 (1 + 7)

2 + 27

Rewrite as

= 2 \cdot 1 + 27

Factor out common term

2

= 2 (1 + 7)

= \frac{2 ( 1 + 7 )}{6}

Cancel the common factor:

2

= \frac{1 + 7}{3}

u = \frac{- ( - 2 ) - 2 7}{2 \cdot 3} : \frac{1 - 7}{3}

\frac{- ( - 2 ) - 2 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 - 2 7}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2 - 2 7}{6}

Factor

2 - 27 : 2 (1 - 7)

2 - 27

Rewrite as

= 2 \cdot 1 - 27

Factor out common term

2

= 2 (1 - 7)

= \frac{2 ( 1 - 7 )}{6}

Cancel the common factor:

2

= \frac{1 - 7}{3}

The solutions to the quadratic equation are:

u = \frac{1 + 7}{3}, u = \frac{1 - 7}{3}

Substitute back

u = sin (θ)

sin (θ) = \frac{1 + 7}{3}, sin (θ) = \frac{1 - 7}{3}

sin (θ) = \frac{1 + 7}{3}, sin (θ) = \frac{1 - 7}{3}

sin (θ) = \frac{1 + 7}{3} :

No Solution

sin (θ) = \frac{1 + 7}{3}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{1 - 7}{3} : θ = arcsin (\frac{1 - 7}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 7}{3}) + 2 πn

sin (θ) = \frac{1 - 7}{3}

Apply trig inverse properties

sin (θ) = \frac{1 - 7}{3}

General solutions for

sin (θ) = \frac{1 - 7}{3}

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

θ = arcsin (\frac{1 - 7}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 7}{3}) + 2 πn

θ = arcsin (\frac{1 - 7}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 7}{3}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1 - 7}{3}) + 2 πn, θ = π + arcsin (- \frac{1 - 7}{3}) + 2 πn

Show solutions in decimal form

θ = - 0.58066 \dots + 2 πn, θ = π + 0.58066 \dots + 2 πn

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tan (2 x - \frac{n}{4}) = - 1, 0^{\circ} < x \leq 36 0^{\circ}

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solvefor x,y+arctan(y/x)=0 solve for

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(θ)-2sin(θ)=2 ?
The general solution for 3sin^2(θ)-2sin(θ)=2 is θ=-0.58066…+2pin,θ=pi+0.58066…+2pin