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

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

Divide both sides by

a; a \neq = 0

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

Apply trig inverse properties

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