How do you solve $cos2x+3cosx-1=0$ ?

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Answer 1 · 2016-09-25T18:13:03

After you substitute $2cos^2(x) - 1$ for $cos(2x)$ , it becomes quadratic equation where $u = cos(x)$ .

$2cos^2(x) - 1 + 3cos(x) - 1 = 0$

$2u^2 + 3u - 2 = 0$

$(u + 2)(2u - 1) = 0$

$u = -2$ is extraneous, because $cos(x)$ cannot equal -2

$cos(x) = 1/2$ is the only possible root.

$x = cos^-1(1/2)$

For the first quadrant:

$x = pi/3$ radians or $60°$

There is a root in the 4th quadrant, $5pi/3$ and, of course, roots at multiples $2pi$ for each.

How do you solve cos2x+3cosx-1=0cos2x+3cosx-1=0?