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

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

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Popular >

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

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Subtract the numbers:

- 2 - 1 = - 3

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 2, c = - 3

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

u_{1} = \frac{- 2 + 4}{2 \cdot 1}, u_{2} = \frac{- 2 - 4}{2 \cdot 1}

u = \frac{- 2 + 4}{2 \cdot 1} : 1

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

Add/Subtract the numbers:

- 2 + 4 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 2 - 4}{2 \cdot 1} : - 3

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

Subtract the numbers:

- 2 - 4 = - 6

= \frac{- 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{2}

Apply the fraction rule:

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

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

The solutions to the quadratic equation are:

u = 1, u = - 3

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - 3

cos (x) = 1, cos (x) = - 3

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) = - 3 :

No Solution

cos (x) = - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

Popular Examples

sin(5x-1)= 4/5

sin (5 x - 1) = \frac{4}{5}

sin^2(x)= 1/36

sin^{2} (x) = \frac{1}{36}

tan(x)=31

tan (x) = 31

sin^4(x)+cos^2(x)=2

sin^{4} (x) + cos^{2} (x) = 2

5tan(x)=15cot(x)

5 tan (x) = 15 cot (x)

Frequently Asked Questions (FAQ)

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