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].

Step-by-step 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)=(1)/(2) \\ x=\cos^(-1)(-1),\text{ }x=\cos^(-1)((1)/(2)) \\ x=\pi,\text{ }x=(\pi)/(3),\text{ }2\pi-(\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=(\pi)/(3),\text{ }(5\pi)/(3) \end{gathered}`

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