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

The general solution for solvefor x,cos^2(x)=2-2sin(x) is x= pi/2+2pin

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

Solution

solve for

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 - 2 sin (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (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 = sin (x)

sin (x) = 1

sin (x) = 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

Popular Examples

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x, 2 \cdot 2.25 \cdot cos (x) = 0

solvefor x,2sin(x)=1 solve for

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cot(x)+2cos(x)=0

cot (x) + 2 cos (x) = 0

2cot(θ)-1=2,cot(θ)-1=0

2 cot (θ) - 1 = 2, cot (θ) - 1 = 0

2cos^2(a)-7cos(a)+3=0

2 cos^{2} (a) - 7 cos (a) + 3 = 0

Frequently Asked Questions (FAQ)

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