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

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

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Popular >

1-cos(x)=sin(x)

Solution

1 - cos (x) = sin (x)

Solution

x = 2 πn, x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

1 - cos (x) = sin (x)

Square both sides

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = cos (x)

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

1 - 1 = 0

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

2 u^{2} - 2 u = 0

Solve with the quadratic formula

2 u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = - 2, c = 0

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

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

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

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

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = cos (x)

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

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

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) = 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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

1 - cos (x) = sin (x)

Remove the ones that don't agree with the equation.

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

1 - cos (x) = sin (x)

plug in

x = 2 π 1

1 - cos (2 π 1) = sin (2 π 1)

Refine

0 = 0

\Rightarrow True

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

1 - cos (x) = sin (x)

plug in

x = \frac{π}{2} + 2 π 1

1 - cos (\frac{π}{2} + 2 π 1) = sin (\frac{π}{2} + 2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

1 - cos (x) = sin (x)

plug in

x = \frac{3 π}{2} + 2 π 1

1 - cos (\frac{3 π}{2} + 2 π 1) = sin (\frac{3 π}{2} + 2 π 1)

Refine

1 = - 1

\Rightarrow False

x = 2 πn, x = \frac{π}{2} + 2 πn

Popular Examples

sin^2(x)=2sin(x)

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

sin(2x)=1+cos(2x)

sin (2 x) = 1 + cos (2 x)

sin(2θ)+sin(4θ)=0

sin (2 θ) + sin (4 θ) = 0

sqrt(3)tan(x)-1=0

3 tan (x) - 1 = 0

cos(4x)=-1/2

cos (4 x) = - \frac{1}{2}

Frequently Asked Questions (FAQ)

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