What is the general solution for 3cos^2(x)-8cos(x)+4=sin^2(x) ?

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

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3cos^2(x)-8cos(x)+4=sin^2(x)

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

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

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

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

No Solution

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

Combine all the solutions

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

cos (x - 4 5^{\circ}) = \frac{- 1}{2}

x, sin (x + \frac{r}{4}) + sin (x - \frac{r}{3}) = 0

cos^{2} (x) = 2 sin^{2} (x)

x, sec (x) - sin (50 o) = 2

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

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