What is the general solution for 4sin^2(x)+2cos(x)+a=3 ?

The general solution for 4sin^2(x)+2cos(x)+a=3 is

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4sin^2(x)+2cos(x)+a=3

4 sin^{2} (x) + 2 cos (x) + a = 3

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

Solution steps

4 sin^{2} (x) + 2 cos (x) + a = 3

Subtract

3

from both sides

4 sin^{2} (x) + 2 cos (x) + a - 3 = 0

Rewrite using trig identities

1 + a + 2 cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

cos (x) = - \frac{- 1 + 4 a + 5}{4}, cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4} : x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

cos (x) = \frac{4 a + 5 + 1}{4} : x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

cos^{5} (x) + cos (x) + 4 cos^{2} (x) = 2

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

\frac{sin ^{3} ( x )}{2 + 2 ( sin ( x ) ) ^{2}} = 1

x, cot^{2} (x) = \frac{1}{3}

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

What is the general solution for 4sin^2(x)+2cos(x)+a=3 ?
The general solution for 4sin^2(x)+2cos(x)+a=3 is