What is the general solution for 2cos^2(x)+3cos(x)=-1,0<= x<= 2pi ?

The general solution for 2cos^2(x)+3cos(x)=-1,0<= x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=pi

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2cos^2(x)+3cos(x)=-1,0<= x<= 2pi

2 cos^{2} (x) + 3 cos (x) = - 1, 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) = - 1, 0 \leq x \leq 2 π

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

9 cos (2 θ) = - 9 cos (θ)

3 sec^{2} (x) - 3 sec (x) = 6

tan (2 θ) = - 3

sin (x) + cos (x) = \frac{2}{5}

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

What is the general solution for 2cos^2(x)+3cos(x)=-1,0<= x<= 2pi ?
The general solution for 2cos^2(x)+3cos(x)=-1,0<= x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=pi