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

Q: What is the general solution for sin^2(x)=2-2cos(x),0<= x<= 2pi ?

The general solution for sin^2(x)=2-2cos(x),0<= x<= 2pi is x=0,x=2pi

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sin^2(x)=2-2cos(x),0<= x<= 2pi

Solution

sin^{2} (x) = 2 - 2 cos (x), 0 \leq x \leq 2 π

Solution

x = 0, x = 2 π

Degrees

x = 0^{\circ}, x = 36 0^{\circ}

Solution steps

sin^{2} (x) = 2 - 2 cos (x), 0 \leq x \leq 2 π

Subtract

2 - 2 cos (x)

from both sides

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

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

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

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

Apply rule

- (- a) = a

= 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 ( - 1 )}

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

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

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

Remove parentheses:

(- a) = - a

= \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 = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

cos (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

cos (x) = 1, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = 2 π

Combine all the solutions

x = 0, x = 2 π

Popular Examples

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- 6 sin (θ) = - 3 + 2 sin (θ)

tan(θ)=(sqrt(7))/(21)

tan (θ) = \frac{7}{21}

(sin^2(x)-2sin(x)+1)/(sin(x)-1)=sin(-1)

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

tan^2(x)=3-2tan(x),0<= x<= 2pi

tan^{2} (x) = 3 - 2 tan (x), 0 \leq x \leq 2 π

4sin(θ)=4cos(θ)

4 sin (θ) = 4 cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)=2-2cos(x),0<= x<= 2pi ?
The general solution for sin^2(x)=2-2cos(x),0<= x<= 2pi is x=0,x=2pi