Solve cos^2x+3cosx+2=0, for cos, then for x.

Question

I do not understand. Please help.

Answer 1

Let's first look at the quadratic function in terms or cosine.

cos²x + 3cosx + 2 = 0

In order to solve for cosx we need to factor the quadratic. To do so it may be easier to rewrite in terms of x instead of cosx.

x² + 3x + 2 = 0

Factoring this results in:

(x+2)(x+1) = 0

Now we can substitute the cosx back in for the x

(cosx + 2)(cos x +1) = 0

This results in 2 solutions of cos x. cosx = -2 and cosx = -1

cosx is undefined for values greater than 1 and less than -1 (Think of the graph of cosine... it only goes up to 1 and only goes down to -1).

So, cosx = -2 has no solution.

cos x = -1 .... where does cosx = -1 or what angle measures give a value of -1 to cosx. On the interval [0,2π] this happens only at π.

So if we are asked to solve on the interval [0,2π], the final answer would be x = π.

But, if does not give us an interval they are likely wanting us to give a general solution which accounts for how the cosine function repeats from -∞ to ∞. In this case we need to add the period of the cosine function multiplied by n. As a refresher the period is the distance on the x-axis over which the cosine function takes to repeat itself. The period of cosx = 2π.

Therefore, the general solution is:

x = π + 2nπ FINAL ANSWER

Hope this explanation helps.

x =

Answer 2

cos²x + 3cosx + 2 = 0

(cosx + 2)(cosx + 1) = 0

cosx = -2 or cosx = -1

Since -1 ≤ cosx ≤1, the equation cosx = -2 has no solution.

So, cosx = -1

x = π + 2kπ = (2k+1)π, k = 0, ±1, ±2, ...