How do you solve $2cos^2(2x)+cos(2x)-1=0$ ?

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Answer 1 · 2016-03-17T05:12:05

$x = pi/6 + kpi or pi/3 + kpi or pi/2 + kpi$ ,

for any integer $k$ .

Let $cos(2x)$ be $y$ . So the equation becomes a standard quadratic

$2y^2 + y - 1 = 0$

Factorize to find the zeroes.

$(2y - 1) * (y + 1) = 0$

This means that $y = 1/2$ or $y = -1$ .

Now change back $y$ to $cos(2x)$ .

So we get

$cos(2x) = 1/2$

or

$cos(2x) = -1$

Next we find the basic angles.

$cos^{-1}(1/2) = pi/3$

$cos^{-1}(-1) = pi$

With that, we solve $cos(2x) = 1/2$ first.

$2x = pi/3 + 2kpi or {2pi}/3 + 2kpi$ ,

for any integer $k$ .

$x = pi/6 + kpi or pi/3 + kpi$

Now, we solve $cos(2x) = -1$ .

$2x = pi + 2kpi$ ,

for any integer $k$ .

$x = pi/2 + kpi$

Combining the solution gives

$x = pi/6 + kpi or pi/3 + kpi or pi/2 + kpi$ ,

for any integer $k$ .

How do you solve 2cos^2(2x)+cos(2x)-1=0 2cos^2(2x)+cos(2x)-1=0 ?