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

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

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3cos(x)-sin^2(x)+3=0

Solution

3 cos (x) - sin^{2} (x) + 3 = 0

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (x) - sin^{2} (x) + 3 = 0

Rewrite using trig identities

3 - sin^{2} (x) + 3 cos (x)

Use the Pythagorean identity:

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

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

= 3 - (1 - cos^{2} (x)) + 3 cos (x)

Simplify

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

3 - (1 - cos^{2} (x)) + 3 cos (x)

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

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

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

= 3 - 1 + cos^{2} (x) + 3 cos (x)

Subtract the numbers:

3 - 1 = 2

= cos^{2} (x) + 3 cos (x) + 2

= cos^{2} (x) + 3 cos (x) + 2

2 + cos^{2} (x) + 3 cos (x) = 0

Solve by substitution

2 + cos^{2} (x) + 3 cos (x) = 0

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 3, c = 2

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

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

3^{2} - 4 \cdot 1 \cdot 2 = 1

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = - 1, u = - 2

Substitute back

u = cos (x)

cos (x) = - 1, cos (x) = - 2

cos (x) = - 1, cos (x) = - 2

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 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}

x = π + 2 πn

x = π + 2 πn

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = π + 2 πn

Popular Examples

solvefor x,cos(x)= 1/4 solve for

x, cos (x) = \frac{1}{4}

cos^2(t)=0

cos^{2} (t) = 0

sin(x)= 18/12

sin (x) = \frac{18}{12}

csc(θ)=sqrt(5)

csc (θ) = 5

cot^2(x)=tan(x/2)

cot^{2} (x) = tan (\frac{x}{2})

Frequently Asked Questions (FAQ)

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