What is the general solution for solvefor x,d^2+3d+2y=4cos^2(x) ?

The general solution for solvefor x,d^2+3d+2y=4cos^2(x) is

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solvefor x,d^2+3d+2y=4cos^2(x)

solve for

x, d^{2} + 3 d + 2 y = 4 cos^{2} (x)

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

Solution steps

d^{2} + 3 d + 2 y = 4 cos^{2} (x)

Switch sides

4 cos^{2} (x) = d^{2} + 3 d + 2 y

Solve by substitution

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}, cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2} : x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2} : x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

Combine all the solutions

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

(1 + tan^{2} (x)) cos^{2} (x) = sec (x)

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

cos^{6} (x) = - cos^{2} (x)

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

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

What is the general solution for solvefor x,d^2+3d+2y=4cos^2(x) ?
The general solution for solvefor x,d^2+3d+2y=4cos^2(x) is