What is the general solution for 1+acos(x)=(a+1)^2 ?

The general solution for 1+acos(x)=(a+1)^2 is x=arccos(a+2)+2pin,x=-arccos(a+2)+2pin

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1+acos(x)=(a+1)^2

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

x = arccos (a + 2) + 2 πn, x = - arccos (a + 2) + 2 πn

Solution steps

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Divide both sides by

a; a \neq = 0

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

Divide both sides by

a; a \neq = 0

\frac{a cos ( x )}{a} = \frac{a ^{2}}{a} + \frac{2 a}{a}; a \neq = 0

Simplify

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

Simplify

\frac{a cos ( x )}{a} : cos (x)

\frac{a cos ( x )}{a}

Cancel the common factor:

a

= cos (x)

Simplify

\frac{a ^{2}}{a} + \frac{2 a}{a} : a + 2

\frac{a ^{2}}{a} + \frac{2 a}{a}

Apply rule

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

= \frac{a ^{2} + 2 a}{a}

Factor

a^{2} + 2 a : a (a + 2)

a^{2} + 2 a

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

a^{2} = aa

= aa + 2 a

Factor out common term

a

= a (a + 2)

= \frac{a ( a + 2 )}{a}

Cancel the common factor:

a

= a + 2

cos (x) = a + 2; a \neq = 0

cos (x) = a + 2; a \neq = 0

cos (x) = a + 2; a \neq = 0

Apply trig inverse properties

cos (x) = a + 2

General solutions for

cos (x) = a + 2

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

x = arccos (a + 2) + 2 πn, x = - arccos (a + 2) + 2 πn

x = arccos (a + 2) + 2 πn, x = - arccos (a + 2) + 2 πn

tan^{4} (x) - 9 = 0

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

2 sin^{2} (w) - sin (w) - 1 = 0

3 sin (x) = 7 cos (x)

8 sin (θ) - 3 = 0

What is the general solution for 1+acos(x)=(a+1)^2 ?
The general solution for 1+acos(x)=(a+1)^2 is x=arccos(a+2)+2pin,x=-arccos(a+2)+2pin