What is the general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 ?

Q: What is the general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 ?

The general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 is No Solution for θ\in\mathbb{R}

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Popular >

2sin(θ)=3cos(θ)-1,0<,θ<360

Solution

2 sin (θ) = 3 cos (θ) - 1, 0 <, θ < 360

Solution

No Solution for θ \in R

Solution steps

2 sin (θ) = 3 cos (θ) - 1, 0, θ < 360

Square both sides

(2 sin (θ))^{2} = (3 cos (θ) - 1)^{2}

Subtract

(3 cos (θ) - 1)^{2}

from both sides

4 sin^{2} (θ) - 9 cos^{2} (θ) + 6 cos (θ) - 1 = 0

Rewrite using trig identities

- 1 + 4 sin^{2} (θ) + 6 cos (θ) - 9 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 1 + 4 (1 - cos^{2} (θ)) + 6 cos (θ) - 9 cos^{2} (θ)

Simplify

- 1 + 4 (1 - cos^{2} (θ)) + 6 cos (θ) - 9 cos^{2} (θ) : 6 cos (θ) - 13 cos^{2} (θ) + 3

- 1 + 4 (1 - cos^{2} (θ)) + 6 cos (θ) - 9 cos^{2} (θ)

Expand

4 (1 - cos^{2} (θ)) : 4 - 4 cos^{2} (θ)

4 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= - 1 + 4 - 4 cos^{2} (θ) + 6 cos (θ) - 9 cos^{2} (θ)

Simplify

- 1 + 4 - 4 cos^{2} (θ) + 6 cos (θ) - 9 cos^{2} (θ) : 6 cos (θ) - 13 cos^{2} (θ) + 3

- 1 + 4 - 4 cos^{2} (θ) + 6 cos (θ) - 9 cos^{2} (θ)

Group like terms

= - 4 cos^{2} (θ) + 6 cos (θ) - 9 cos^{2} (θ) - 1 + 4

Add similar elements:

- 4 cos^{2} (θ) - 9 cos^{2} (θ) = - 13 cos^{2} (θ)

= - 13 cos^{2} (θ) + 6 cos (θ) - 1 + 4

Add/Subtract the numbers:

- 1 + 4 = 3

= 6 cos (θ) - 13 cos^{2} (θ) + 3

= 6 cos (θ) - 13 cos^{2} (θ) + 3

= 6 cos (θ) - 13 cos^{2} (θ) + 3

3 - 13 cos^{2} (θ) + 6 cos (θ) = 0

Solve by substitution

3 - 13 cos^{2} (θ) + 6 cos (θ) = 0

Let:

cos (θ) = u

3 - 13 u^{2} + 6 u = 0

3 - 13 u^{2} + 6 u = 0 : u = - \frac{- 3 + 4 3}{13}, u = \frac{3 + 4 3}{13}

3 - 13 u^{2} + 6 u = 0

Write in the standard form

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

- 13 u^{2} + 6 u + 3 = 0

Solve with the quadratic formula

- 13 u^{2} + 6 u + 3 = 0

Quadratic Equation Formula:

For

a = - 13, b = 6, c = 3

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 13 ) \cdot 3}{2 ( - 13 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 13 ) \cdot 3}{2 ( - 13 )}

6^{2} - 4 (- 13) \cdot 3 = 83

6^{2} - 4 (- 13) \cdot 3

Apply rule

- (- a) = a

= 6^{2} + 4 \cdot 13 \cdot 3

Multiply the numbers:

4 \cdot 13 \cdot 3 = 156

= 6^{2} + 156

6^{2} = 36

= 36 + 156

Add the numbers:

36 + 156 = 192

= 192

Prime factorization of

192 : 2^{6} \cdot 3

192

192

divides by

2192 = 96 \cdot 2

= 2 \cdot 96

96

divides by

296 = 48 \cdot 2

= 2 \cdot 2 \cdot 48

48

divides by

248 = 24 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

= 2^{6} \cdot 3

= 2^{6} \cdot 3

Apply radical rule:

= 3 2^{6}

Apply radical rule:

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 3

Refine

= 83

u_{1, 2} = \frac{- 6 \pm 8 3}{2 ( - 13 )}

Separate the solutions

u_{1} = \frac{- 6 + 8 3}{2 ( - 13 )}, u_{2} = \frac{- 6 - 8 3}{2 ( - 13 )}

u = \frac{- 6 + 8 3}{2 ( - 13 )} : - \frac{- 3 + 4 3}{13}

\frac{- 6 + 8 3}{2 ( - 13 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 8 3}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{- 6 + 8 3}{- 26}

Apply the fraction rule:

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

= - \frac{- 6 + 8 3}{26}

Cancel

\frac{- 6 + 8 3}{26} : \frac{4 3 - 3}{13}

\frac{- 6 + 8 3}{26}

Factor

- 6 + 83 : 2 (- 3 + 43)

- 6 + 83

Rewrite as

= - 2 \cdot 3 + 2 \cdot 43

Factor out common term

2

= 2 (- 3 + 43)

= \frac{2 ( - 3 + 4 3 )}{26}

Cancel the common factor:

2

= \frac{- 3 + 4 3}{13}

= - \frac{4 3 - 3}{13}

= - \frac{- 3 + 4 3}{13}

u = \frac{- 6 - 8 3}{2 ( - 13 )} : \frac{3 + 4 3}{13}

\frac{- 6 - 8 3}{2 ( - 13 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 8 3}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{- 6 - 8 3}{- 26}

Apply the fraction rule:

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

- 6 - 83 = - (6 + 83)

= \frac{6 + 8 3}{26}

Factor

6 + 83 : 2 (3 + 43)

6 + 83

Rewrite as

= 2 \cdot 3 + 2 \cdot 43

Factor out common term

2

= 2 (3 + 43)

= \frac{2 ( 3 + 4 3 )}{26}

Cancel the common factor:

2

= \frac{3 + 4 3}{13}

The solutions to the quadratic equation are:

u = - \frac{- 3 + 4 3}{13}, u = \frac{3 + 4 3}{13}

Substitute back

u = cos (θ)

cos (θ) = - \frac{- 3 + 4 3}{13}, cos (θ) = \frac{3 + 4 3}{13}

cos (θ) = - \frac{- 3 + 4 3}{13}, cos (θ) = \frac{3 + 4 3}{13}

cos (θ) = - \frac{- 3 + 4 3}{13}, 0 :

No Solution

cos (θ) = - \frac{- 3 + 4 3}{13}, 0

Apply trig inverse properties

cos (θ) = - \frac{- 3 + 4 3}{13}

General solutions for

cos (θ) = - \frac{- 3 + 4 3}{13}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{- 3 + 4 3}{13}) + 2 πn, θ = - arccos (- \frac{- 3 + 4 3}{13}) + 2 πn

θ = arccos (- \frac{- 3 + 4 3}{13}) + 2 πn, θ = - arccos (- \frac{- 3 + 4 3}{13}) + 2 πn

Solutions for the range

0

No Solution

cos (θ) = \frac{3 + 4 3}{13}, 0 :

No Solution

cos (θ) = \frac{3 + 4 3}{13}, 0

Apply trig inverse properties

cos (θ) = \frac{3 + 4 3}{13}

General solutions for

cos (θ) = \frac{3 + 4 3}{13}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{3 + 4 3}{13}) + 2 πn, θ = 2 π - arccos (\frac{3 + 4 3}{13}) + 2 πn

θ = arccos (\frac{3 + 4 3}{13}) + 2 πn, θ = 2 π - arccos (\frac{3 + 4 3}{13}) + 2 πn

Solutions for the range

0

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 sin (θ) = 3 cos (θ) - 1

Remove the ones that don't agree with the equation.

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 ?
The general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 is No Solution for θ\in\mathbb{R}