What is the general solution for solvefor A,y=-6cos(A)-2sin^2(A) ?

The general solution for solvefor A,y=-6cos(A)-2sin^2(A) is

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solvefor A,y=-6cos(A)-2sin^2(A)

solve for

A, y = - 6 cos (A) - 2 sin^{2} (A)

A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

Solution steps

y = - 6 cos (A) - 2 sin^{2} (A)

Switch sides

- 6 cos (A) - 2 sin^{2} (A) = y

Subtract

y

from both sides

- 6 cos (A) - 2 sin^{2} (A) - y = 0

Rewrite using trig identities

- y - (1 - cos^{2} (A)) \cdot 2 - 6 cos (A) = 0

Solve by substitution

cos (A) = \frac{3 + 2 y + 13}{2}, cos (A) = \frac{3 - 2 y + 13}{2}

cos (A) = \frac{3 + 2 y + 13}{2} : A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn

cos (A) = \frac{3 - 2 y + 13}{2} : A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

Combine all the solutions

A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

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

sin^{4} (x) = cos^{4} (x)

10. 5^{2} = 6. 7^{2} + 7. 8^{2} - 2 \cdot 6.7 \cdot 7.8 \cdot cos (x)

3 sinh (x) - cosh (x) = 1

tan (4 5^{\circ} + \frac{x}{3}) = 0.58

What is the general solution for solvefor A,y=-6cos(A)-2sin^2(A) ?
The general solution for solvefor A,y=-6cos(A)-2sin^2(A) is