What is the general solution for -8cos^2(x)+1=4sin^2(x)-8cos(x) ?

The general solution for -8cos^2(x)+1=4sin^2(x)-8cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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-8cos^2(x)+1=4sin^2(x)-8cos(x)

- 8 cos^{2} (x) + 1 = 4 sin^{2} (x) - 8 cos (x)

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

- 8 cos^{2} (x) + 1 = 4 sin^{2} (x) - 8 cos (x)

Subtract

4 sin^{2} (x) - 8 cos (x)

from both sides

- 8 cos^{2} (x) + 1 - 4 sin^{2} (x) + 8 cos (x) = 0

Rewrite using trig identities

- 3 - 4 cos^{2} (x) + 8 cos (x) = 0

Solve by substitution

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{3}{2} :

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

\frac{1 + csc ( y )}{cot ( y ) + cos ( y )} = csc (y)

2 = sin (10 x)

cos (2 x) = \frac{( 2 k - 8 )}{9}

x, h = tan (x - \frac{π}{4})

cos^{3} (x) + sin (x) = 0

What is the general solution for -8cos^2(x)+1=4sin^2(x)-8cos(x) ?
The general solution for -8cos^2(x)+1=4sin^2(x)-8cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin