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

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

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sin^2(x)+3cos(x)-1=2

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Subtract

2

from both sides

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

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

3^{2} - 4 (- 1) (- 2) = 1

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

Apply rule

- (- a) = a

= 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 ( - 1 )}

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

= \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 ( - 1 )} : 2

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

Remove parentheses:

(- a) = - a

= \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 = 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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

Popular Examples

cot^2(θ)-3csc(θ)+3=0

cot^{2} (θ) - 3 csc (θ) + 3 = 0

solvefor x,arctan(x)+arctan(y)= pi/2 solve for

x, arctan (x) + arctan (y) = \frac{π}{2}

2sin(x)=2sin(2x)

2 sin (x) = 2 sin (2 x)

3cos(2x)-1=0

3 cos (2 x) - 1 = 0

2cos^2(x)=2

2 cos^{2} (x) = 2

Frequently Asked Questions (FAQ)

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