Question

Solve for all values of a on the interval [0, 2].2 cos²x+3 = -cos x + 4

Answer 1

Given

The equation 2 cos²x + 3 = -cos x + 4.

To solve for all values of x on the interval [0, 2π].

Explanation:

It is given that,

The equation 2 cos²x + 3 = -cos x + 4.

That implies,

$\begin{gathered} 2\cos²x+3=-\cos x+4 \\ 2\cos^2x+\cos x+3-4=0 \\ 2\cos^2x+\cos x-1=0 \end{gathered}$

Substitute cosx = a.

Then,

$\begin{gathered} 2a^2+a-1=0 \\ 2a^2+2a-a-1=0 \\ 2a(a+1)-(a+1)=0 \\ (a+1)(2a-1)=0 \\ a+1=0,\text{ }2a-1=0 \end{gathered}$

That implies,

$\begin{gathered} \cos x+1=0,\text{ }2\cos x-1=0 \\ \cos x=-1,\text{ }2\cos x=1 \\ \cos(x)=-1,\text{ }\cos(x)=\frac{1}{2} \\ x=\cos^{-1}(-1),\text{ }x=\cos^{-1}(\frac{1}{2}) \\ x=\pi,\text{ }x=\frac{\pi}{3},\text{ }2\pi-\frac{\pi}{3}[since\text{ }cosine\text{ }is\text{ }positive\text{ }in\text{ }the\text{ }1^{st}\text{ \& }3^{rd}\text{ }quadrant] \\ x=\pi,\text{ }x=\frac{\pi}{3},\text{ }\frac{5\pi}{3} \end{gathered}$

Hence, the solution is, x=π, π/3, 5π/3.