Don't get scared off by the fact we're doing trig functions! Factor as you normally would. Because our middle term is negative (\(\displaystyle -2\cos x\)), we know that the signs inside of our parentheses will be negative. 

This means that \(\displaystyle 1-2\cos x+\cos^2 x\) can be factored to \(\displaystyle (1-\cos x)(1-\cos x)\) or \(\displaystyle (1-\cos x)^2\).

The fastest way to solve this equation is to simply try the three answers.  Plugging in \(\displaystyle \frac{\pi}{4}\) gives

\(\displaystyle tan^2(\frac{\pi}{4})-tan(\frac{\pi}{4})=(1)^2-1=1-1=0\)

Our first choice is valid.

Plugging in \(\displaystyle \frac{\pi}{2}\) gives

\(\displaystyle tan^2(\frac{\pi}{2})-tan(\frac{\pi}{2})\)

However, since \(\displaystyle tan(\frac{\pi}{}2)\) is undefined, this cannot be a valid answer.

Finally, plugging in \(\displaystyle \frac{3\pi}{4}\) gives

\(\displaystyle tan^2(\frac{3\pi}{4})-tan(\frac{3\pi}{4})=(-1)^2-(-1)=1+1=2\)

Therefore, our third answer choice is not correct, meaning only 1 is correct.

\(\displaystyle f(x)=\sin^2(x)-\sin(2x)+\cos^2(x)=1-2\sin(x)cos(x)\)

\(\displaystyle =1-2\sin(x)cos(x)=0\)

Therefore, 

\(\displaystyle 1=2sin(x)cos(x)\)

\(\displaystyle \frac{1}{2}=sin(x)cos(x)\)

and that only happens once in the given interval, at \(\displaystyle \frac{\pi}{4}\), or 45 degrees.

\(\displaystyle \sin\bigg(\frac{\pi}{4}\bigg)\cos\bigg(\frac{\pi}{4}\bigg)=\frac{1}{\sqrt2}\cdot \frac{1}{\sqrt2}=\frac{1}{2}\)

We have \(\displaystyle cos^2x+sin^2x=1\).

Now since \(\displaystyle 1+2cos(x)sin(x)=cos^2(x)+sin^2(x)+2sin(x)cos(x)\)

This last expression can be written as :

\(\displaystyle (cos(x)+sin(x))^2\).

This shows the required result.

We know that we can write

\(\displaystyle a^3-1\) in the following form

\(\displaystyle (a^3-1)=(a-1)(a^2+a+1)\).

Now taking \(\displaystyle a=sinx\),

we have:

\(\displaystyle sin^3 x-1=(sinx-1)(sin^2 x+sinx+1)\).

This is the result that we need.

First we see that :

\(\displaystyle a^4-b^4=(a^2+b^2)(a^2-b^2)\).

Now letting \(\displaystyle a=cosx ,b=sinx\)

we have

\(\displaystyle cos^4x-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)\)

We know that :

\(\displaystyle cos^2x+sin^2x=1\) and we are given that

\(\displaystyle cos^2x-sin^2x=cos2x\), this gives

\(\displaystyle cos^4x-sin^4x=cos2x\)

Note first that:

\(\displaystyle a^8-b^8=(a^4-b^4)(a^4+b^4)\) and :

\(\displaystyle (a^4-b^4)=(a^2+b^2)(a^2 - b^2)\).

Now taking \(\displaystyle a=cosx \quad b=sinx\). We have

\(\displaystyle cos^8x-sin^8x=(cos^4x+sin^4x)(cos^4-sin^4x)\).

Since \(\displaystyle cos^4x-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)\) and \(\displaystyle cos^2x+sin^2x=1\).

We therefore have :

\(\displaystyle cos^8x-sin^8x=(cos^4x+sin^4x)(cos^2x-sin^2x)\)

Letting \(\displaystyle sin^n(x)=X\), we have the equivalent expression:

\(\displaystyle X^2+2X+3\).

We cant factor \(\displaystyle X^2+2X+3\) since \(\displaystyle X^2+2X+3=(x+1)^2+2 >0 \forall \quadx\in\mathbb{R}\).

This shows that we cannot factor the above expression.

We first note that we have:

\(\displaystyle 1-a^{11}=(1-a)(a^{10}+a^9+...a+1)\)

Then taking \(\displaystyle a=cos(x)\), we have the result.

\(\displaystyle (1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)\)

First of all we know that :

\(\displaystyle cos^2x+sin^2x=1\) and this gives:

\(\displaystyle 1-cos^2{x}=sin^{2}x\).

Now we need to see that: \(\displaystyle cos^{2n}x-cos^{2n+2} x\) can be written as

\(\displaystyle cos^{2n}x(1-cos^2x)\) and since \(\displaystyle 1-cos^2{x}=sin^{2}x\)

we have then:

\(\displaystyle cos^{2n}x(1-cos^2x)=cos^{2n}xsin^2x\).