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

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

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Popular >

3sin^2(θ)-cos(θ)+1=0

Solution

3 sin^{2} (θ) - cos (θ) + 1 = 0

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin^{2} (θ) - cos (θ) + 1 = 0

Rewrite using trig identities

1 - cos (θ) + 3 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 - cos (θ) + 3 (1 - cos^{2} (θ))

Simplify

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

1 - cos (θ) + 3 (1 - cos^{2} (θ))

Expand

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

3 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (θ)

= 3 \cdot 1 - 3 cos^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 cos^{2} (θ)

= 1 - cos (θ) + 3 - 3 cos^{2} (θ)

Simplify

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

1 - cos (θ) + 3 - 3 cos^{2} (θ)

Group like terms

= - cos (θ) - 3 cos^{2} (θ) + 1 + 3

Add the numbers:

1 + 3 = 4

= - 3 cos^{2} (θ) - cos (θ) + 4

= - 3 cos^{2} (θ) - cos (θ) + 4

= - 3 cos^{2} (θ) - cos (θ) + 4

4 - cos (θ) - 3 cos^{2} (θ) = 0

Solve by substitution

4 - cos (θ) - 3 cos^{2} (θ) = 0

Let:

cos (θ) = u

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

4 - u - 3 u^{2} = 0 : u = - \frac{4}{3}, u = 1

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

Write in the standard form

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

- 3 u^{2} - u + 4 = 0

Solve with the quadratic formula

- 3 u^{2} - u + 4 = 0

Quadratic Equation Formula:

For

a = - 3, b = - 1, c = 4

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

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

(- 1)^{2} - 4 (- 3) \cdot 4 = 7

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 3 \cdot 4

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 3 \cdot 4 = 48

4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 7}{2 ( - 3 )} : - \frac{4}{3}

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

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + 7}{- 2 \cdot 3}

Add the numbers:

1 + 7 = 8

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{8}{- 6}

Apply the fraction rule:

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

= - \frac{8}{6}

Cancel the common factor:

2

= - \frac{4}{3}

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

\frac{- ( - 1 ) - 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - 7}{- 2 \cdot 3}

Subtract the numbers:

1 - 7 = - 6

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{- 6}

Apply the fraction rule:

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

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{4}{3}, u = 1

Substitute back

u = cos (θ)

cos (θ) = - \frac{4}{3}, cos (θ) = 1

cos (θ) = - \frac{4}{3}, cos (θ) = 1

cos (θ) = - \frac{4}{3} :

No Solution

cos (θ) = - \frac{4}{3}

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = 2 πn

Popular Examples

2-2cos^2(x)=0

2 - 2 cos^{2} (x) = 0

3csc^3(x)-4csc(x)=0

3 csc^{3} (x) - 4 csc (x) = 0

tan(θ)=(6.06)/(-33.5)

tan (θ) = \frac{6.06}{- 33.5}

csc(θ)=(sqrt(13))/2 ,sec(θ)=(sqrt(13))/3

csc (θ) = \frac{13}{2}, sec (θ) = \frac{13}{3}

4cos^2(x)+cos(2x)-7cos(x)=-2,0<= x<= 360

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2, 0^{\circ} \leq x \leq 36 0^{\circ}

Frequently Asked Questions (FAQ)

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