When Is Cosine Squared 1?

$\sin^2(x) + \cos^2(x)\equiv1\Rightarrow\sin^2(x) = 0\Rightarrow\sin(x) = 0 \therefore\theta\ = n\pi$ (note: I used the sine graph to determine the value of $\theta$ )

$\cos^{2}\ \theta=1$

$\cos\ \theta=\pm \sqrt{1}$

$\cos\ \theta=\pm{1}$

$\theta'= \arccos\ (1)$

$T=\{\exists\ \theta' \mid \theta'=2n\pi,\ n \in \mathbb{N}\}$

$\theta''=\arccos\ (-1)$

$S=\{\exists\ \theta'' \mid \theta''=(2n+1)\pi,\ n \in \mathbb{N}\}$

$A=\{\exists \ \theta \mid \theta \in (S \cup T)\}$

$\boxed{\theta=n \pi}$

The only solutions to the equation $x^2 = 1$ are when $x = 1$ and $x = -1$ ; by substitution, this means we want when $\cos(\theta) = \pm 1 .$

This occurs only at the extrema of the cosine; in the domain from $[0, 2\pi )$ they are at $0$ and $\pi .$

Including all coterminal angles obtained via adding or subtracting $2\pi$ , we want the sequence $... -2\pi, -\pi, 0, \pi, 2\pi, ...$ which can be expressed by $\theta = n\pi .$

• $\displaystyle \cos ^{ 2 }{ \theta } =1\\ \Rightarrow \cos ^{ 2 }{ \theta } -1=0\\ \Rightarrow \left( \cos { \theta } +1 \right) \left( \cos { \theta } -1 \right) =0$

• For $\cos { \theta } -1=0$ , $\cos { \theta } =1$ . The smallest positive angle, whose cosine is $1$ , is $0$ . The general expression of all the angles which have this cosine is $2n\pi \pm 0$ , i.e., even multiples of $\pi$

• Similarly, the smallest positive angle satisfying $\cos { \theta } +1=0$ is $\theta =\pi$ . The general expression of all the (solution) angles is $2n\pi \pm \pi$ , i.e., odd multiples of $\pi$

• The option $\theta =n\pi$ is union of general solutions (angles) of both the above cases, i.e. integral multiple of $\pi$ , Hence the answer.

$cos(n\pi)=(-1)^n$

$cos^2(n\pi)=(-1)^{2n}$

$cos^2(n\pi)=1$

$Therefore$

$\theta=n\pi$

$(\cos \theta)^{2} = 1$

$(1+ \cos 2\theta)/2 = 1$

$\cos 2 \theta = 1$

$\theta = n \pi$

I have added a written solution for this problem above this one for ease.

Cos valus npai equals1

for all Cos $ =1 ...we know possible $=n.Pie