Answer:
[tex]\frac{2\pi }{3} \ ,\ \pi \ ,\ \frac{4\pi }{3}[/tex]
Step-by-step explanation:
suppose cos(x) = y
Then
[tex]2\cos^{2} \left( x\right) +3\cos x+1=0\Longleftrightarrow \begin{cases}2y^{2}+3y+1=0&\\ y=\cos \left( x\right) \ \ \text{and} &0\leqslant x\leqslant 2\pi \end{cases}[/tex]
2y² + 3y + 1 = 0 ⇒ y = -1 or y = -1/2 (because 2-3+1=0)
y = -1 ⇔ cosx = -1 ⇒ x = π (because x∈[0,2π] )
y = -1/2 ⇔ cosx = -1/2 ⇒ x = 2π/3 or x = 4π/3
Final answer :
x = π or x = 2π/3 or x = 4π/3
