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

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

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

u - u^{2} = 0

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

u - u^{2} = 0

Write in the standard form

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

- u^{2} + u = 0

Solve with the quadratic formula

- u^{2} + u = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 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

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = 1

cos (x) = 0, cos (x) = 1

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn

Popular Examples

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tan (\frac{2 x}{3}) = 0

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3 cos^{2} (A) + 3 cos (A) = 0

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cos(3x)(2cos(x)+1)=0

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3csc(θ)-1=0

3 csc (θ) - 1 = 0

Frequently Asked Questions (FAQ)

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