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

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

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

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Subtract

2 cos (x) - 2

from both sides

- 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) + 1

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 = 1

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

u^{2} - 2 u + 1 = 0

Solve with the quadratic formula

u^{2} - 2 u + 1 = 0

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

u = \frac{- ( - 2 )}{2 \cdot 1}

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

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

Combine all the solutions

x = 2 πn

Popular Examples

3sin(2θ)-5sin(θ)=0

3 sin (2 θ) - 5 sin (θ) = 0

cos(2x)=cos(x),0<= x<= 2pi

cos (2 x) = cos (x), 0 \leq x \leq 2 π

tan(x)=(sqrt(3))/2

tan (x) = \frac{3}{2}

cos(x)= 7/13

cos (x) = \frac{7}{13}

cos(pi/2 x)=0

cos (\frac{π}{2} x) = 0

Frequently Asked Questions (FAQ)

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