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

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

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

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

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

Solution steps

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

Subtract

- 3 cos (x)

from both sides

2 + cos (2 x) + 3 cos (x) = 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 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

Combine all the solutions

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

sec (x + 1 0^{\circ}) = 5.759

cos (3 x + \frac{π}{4}) = \frac{3}{2}

- sin (4 x) = sin (8 x)

tan (θ) = \frac{3 3}{3}

cos (2 x) = 1 - sin^{2} (2 x), 0 \leq x \leq 2 π

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