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Popular >

(sin^2(x))/2+c=cos(x)c+c

Solution

\frac{sin ^{2} ( x )}{2} + c = cos (x) c + c

Solution

x = arccos (- c - c^{2} + 1) + 2 πn, x = - arccos (- c - c^{2} + 1) + 2 πn, x = arccos (- c + c^{2} + 1) + 2 πn, x = - arccos (- c + c^{2} + 1) + 2 πn

Solution steps

\frac{sin ^{2} ( x )}{2} + c = cos (x) c + c

Subtract

cos (x) c + c

from both sides

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

Simplify

- c cos (x) + \frac{sin ^{2} ( x )}{2} : \frac{- 2 c cos ( x ) + sin ^{2} ( x )}{2}

- c cos (x) + \frac{sin ^{2} ( x )}{2}

Convert element to fraction:

c cos (x) = \frac{c cos ( x ) 2}{2}

= - \frac{c cos ( x ) \cdot 2}{2} + \frac{sin ^{2} ( x )}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- c cos ( x ) \cdot 2 + sin ^{2} ( x )}{2}

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

(- 2 c)^{2} - 4 (- 1) \cdot 1 : 2 c^{2} + 1

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

Apply rule

- (- a) = a

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

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

(- 2 c)^{2}

Apply exponent rule:

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

n

is even

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

= (2 c)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} c^{2}

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 2^{2} c^{2} + 4

Factor

2^{2} c^{2} + 4 : 4 (c^{2} + 1)

2^{2} c^{2} + 4

Rewrite as

= 4 c^{2} + 4 \cdot 1

Factor out common term

4

= 4 (c^{2} + 1)

= 4 (c^{2} + 1)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 4 c^{2} + 1

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 c^{2} + 1

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 c + 2 c ^{2} + 1}{- 2}

Apply the fraction rule:

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

= - \frac{2 c + 2 c ^{2} + 1}{2}

Cancel

\frac{2 c + 2 c ^{2} + 1}{2} : c + c^{2} + 1

\frac{2 c + 2 c ^{2} + 1}{2}

Factor out common term

2

= \frac{2 ( c + c ^{2} + 1 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= c + c^{2} + 1

= - (c + c^{2} + 1)

Distribute parentheses

= - (c) - (c^{2} + 1)

Apply minus-plus rules

+ (- a) = - a

= - c - c^{2} + 1

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 c - 2 c ^{2} + 1}{- 2}

Apply the fraction rule:

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

= - \frac{2 c - 2 c ^{2} + 1}{2}

Cancel

\frac{2 c - 2 c ^{2} + 1}{2} : c - c^{2} + 1

\frac{2 c - 2 c ^{2} + 1}{2}

Factor out common term

2

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

Divide the numbers:

\frac{2}{2} = 1

= c - c^{2} + 1

= - (c - c^{2} + 1)

Distribute parentheses

= - (c) - (- c^{2} + 1)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - c + c^{2} + 1

The solutions to the quadratic equation are:

u = - c - c^{2} + 1, u = - c + c^{2} + 1

Substitute back

u = cos (x)

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

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

cos (x) = - c - c^{2} + 1 : x = arccos (- c - c^{2} + 1) + 2 πn, x = - arccos (- c - c^{2} + 1) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (- c - c^{2} + 1) + 2 πn, x = - arccos (- c - c^{2} + 1) + 2 πn

x = arccos (- c - c^{2} + 1) + 2 πn, x = - arccos (- c - c^{2} + 1) + 2 πn

cos (x) = - c + c^{2} + 1 : x = arccos (- c + c^{2} + 1) + 2 πn, x = - arccos (- c + c^{2} + 1) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (- c + c^{2} + 1) + 2 πn, x = - arccos (- c + c^{2} + 1) + 2 πn

x = arccos (- c + c^{2} + 1) + 2 πn, x = - arccos (- c + c^{2} + 1) + 2 πn

Combine all the solutions

x = arccos (- c - c^{2} + 1) + 2 πn, x = - arccos (- c - c^{2} + 1) + 2 πn, x = arccos (- c + c^{2} + 1) + 2 πn, x = - arccos (- c + c^{2} + 1) + 2 πn

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