What is the general solution for cos(2x)=-3cos(x)-2,0<x<= 2pi ?

The general solution for cos(2x)=-3cos(x)-2,0<x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=pi

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cos(2x)=-3cos(x)-2,0<x<= 2pi

cos (2 x) = - 3 cos (x) - 2, 0 < x \leq 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = π

Solution steps

cos (2 x) = - 3 cos (x) - 2, 0 < x \leq 2 π

Subtract

- 3 cos (x) - 2

from both sides

cos (2 x) + 3 cos (x) + 2 = 0

Rewrite using trig identities

1 + 2 cos^{2} (x) + 3 cos (x) = 0

Solve by substitution

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

cos (x) = - \frac{1}{2}, 0 < x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = - 1, 0 < x \leq 2 π : x = π

Combine all the solutions

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = π

cos (θ) = - \frac{7}{6}

csc (θ) = 5, sin (θ) = x

tan (θ) = 0.6

tan (θ) = 0.9

cos (W) = \frac{17 0 ^{2} + 17 3 ^{2} - 12 5 ^{2}}{2 ( 170 ) ( 173 )}

What is the general solution for cos(2x)=-3cos(x)-2,0<x<= 2pi ?
The general solution for cos(2x)=-3cos(x)-2,0<x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=pi