What is the general solution for 3cos^2(θ)+2sin(θ)-1=0 ?

The general solution for 3cos^2(θ)+2sin(θ)-1=0 is θ=-0.58066…+2pin,θ=pi+0.58066…+2pin

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Popular >

3cos^2(θ)+2sin(θ)-1=0

Solution

3 cos^{2} (θ) + 2 sin (θ) - 1 = 0

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 cos^{2} (θ) + 2 sin (θ) - 1 = 0

Rewrite using trig identities

- 1 + 2 sin (θ) + 3 cos^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

- 1 + 2 sin (θ) + 3 (1 - sin^{2} (θ))

Expand

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

3 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (θ)

= 3 \cdot 1 - 3 sin^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (θ)

= - 1 + 2 sin (θ) + 3 - 3 sin^{2} (θ)

Simplify

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

- 1 + 2 sin (θ) + 3 - 3 sin^{2} (θ)

Group like terms

= 2 sin (θ) - 3 sin^{2} (θ) - 1 + 3

Add/Subtract the numbers:

- 1 + 3 = 2

= 2 sin (θ) - 3 sin^{2} (θ) + 2

= 2 sin (θ) - 3 sin^{2} (θ) + 2

= 2 sin (θ) - 3 sin^{2} (θ) + 2

2 + 2 sin (θ) - 3 sin^{2} (θ) = 0

Solve by substitution

2 + 2 sin (θ) - 3 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

2 + 2 u - 3 u^{2} = 0 : u = - \frac{- 1 + 7}{3}, u = \frac{1 + 7}{3}

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

Write in the standard form

a x^{2} + b x + c = 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 ( - 3 ) \cdot 2}{2 ( - 3 )}

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

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

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

Apply rule

- (- a) = a

= 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 ( - 3 )}

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 3 = 6

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

Apply the fraction rule:

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

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

Cancel

\frac{- 2 + 2 7}{6} : \frac{7 - 1}{3}

\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}

= - \frac{7 - 1}{3}

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 3 = 6

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

Apply the fraction rule:

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

- 2 - 27 = - (2 + 27)

= \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} : 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

sin (θ) = \frac{1 + 7}{3} :

No Solution

sin (θ) = \frac{1 + 7}{3}

- 1 \leq sin (x) \leq 1

No Solution

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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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(θ)+2sin(θ)-1=0 ?
The general solution for 3cos^2(θ)+2sin(θ)-1=0 is θ=-0.58066…+2pin,θ=pi+0.58066…+2pin