$\cos^2 x = 1$
How do you solve trig equations with a power?
Unsure what to do with the square?
I get this
- $\frac{1+\cos2x}2 =1$
- $\cos2x =1$
- $2x=2n\pi\pm0$
- $x=n\pi$
but the answer says $\pm n\pi$
$\endgroup$ 16 Answers
$\begingroup$Using the identity: $cos^2x + sin^2x = 1$. So $sin^2x = 0$, and $sinx = 0$, so $x = n\pi$
$\endgroup$ $\begingroup$$\cos^2x-1=0$
$(\cos x -1)(\cos x +1)=0$
$\cos x =1 \ \ \ \ $ or $\ \ \ \cos x = -1$
$x=0+2k \pi \ \ $ or $ \ \ \ x= \pi + 2 k \pi$
$ k =0, \pm 1, \pm 2, \ldots$
$\endgroup$ $\begingroup$$\cos^2 x = 1$
Just square root both sides to get:
$\cos x = \pm 1$
So any angles that have a cosine of $-1$ and any angles that have a cosine of $1$ will satisfy this equation.
$x = \{0+2\pi k, \pi+2 \pi k\},$ where $k$ is any integer.
So this solution set will generate all of the angles that have a cosine of either $-1$ or $1$ because when you add $2\pi$ to an angle you just get an angle that's in the same place which has the same cosine.
$\endgroup$ 3 $\begingroup$Use the identity $$\cos^2(x)=\frac{1+\cos(2x)}{2}.$$ You should then be able to solve this for $x$ by way of the inverse.
$\endgroup$ 7 $\begingroup$Hint:
$\cos^2(x) = \frac 1 2(\cos(2x) + 1)$
$\endgroup$ 2 $\begingroup$Hint
$$\cos^2(x) = \frac{1}{2} + \frac{1}{2}\cos(2x)$$
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