Use the trigonometric identities to switch sec into terms of tan:

\(\displaystyle tan^2\theta + sec^2\theta = tan^2\theta + (tan^2 \theta + 1) = 2tan^2\theta + 1 = 1\)

hence,

\(\displaystyle 2tan^2\theta = 0\)

So we have \(\displaystyle \tan \theta = 0\), making \(\displaystyle \theta= n \pi\)

Therefore the solution is \(\displaystyle \theta = n \pi\) for n being any integer.

We begin by setting the right side of the equation equal to 0.

\(\displaystyle \small 2sin^2(x)-3sin(x)+1=0\)

The equation might be easier to factor using the following substitution.

\(\displaystyle \small w=sin(x)\)

This gives the following

\(\displaystyle \small 2w^2-3w+1=0\)

This can be factored as follows

\(\displaystyle \small \small (2w-1)(w-1)=0\)

Therefore

\(\displaystyle \small w=\frac{1}{2}\;or\;w=1\)

Replacing our substitution therefore gives

\(\displaystyle \small sin(x)=\frac{1}{2}\;or\;sin(x)=1\)

Within our designated domain, we get three answers between our two equations.

\(\displaystyle \small sin(x)=\frac{1}{2}\)       when \(\displaystyle \small x=\frac{\pi}{6}\;or\; \frac{5\pi}{6}\)

\(\displaystyle \small sin(x)=1\)        when \(\displaystyle \small x=\frac{\pi}{2}\)

Therefore, the only choice that isn't correct is \(\displaystyle \small \frac{11\pi}{6}\)

Begin by isolating the tangent side of the equation:

\(\displaystyle tan(5x)+17=-55\)

\(\displaystyle tan(5x)=-72\)

Next, take the inverse tangent of both sides:

\(\displaystyle 5x=tan^{-1} (-72)\)

Divide by five to get the final answer:

\(\displaystyle x=\frac{tan^{-1} (-72)}{5}=-17.84^{\circ}\)

There are two ways to solve this problem. The first involves two trigonometric identities:

\(\displaystyle \tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13\\ \implies\tan^2\theta(\sin^2\theta+\cos^2\theta)+1=13\\ \implies\tan^2\theta+1=13\implies\sec^2\theta=13\\ \implies\sec\theta=\sqrt13\implies\cos\theta=\frac{\sqrt13}{13}\\ \implies \theta=\cos^{-1}\frac{\sqrt13}{13}}\approx73.898}\)

The second method allows us to only use the first trigonometric identity:

\(\displaystyle \tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13\\ \implies\tan^2\theta(\sin^2\theta+\cos^2\theta)+1=13\\ \implies\tan^2\theta+1=13\implies\tan^2\theta=12\\ \implies\tan\theta=2\sqrt3\implies\theta=\tan^{-1}2\sqrt3\approx73.898}\)

The simplest way to solve this problem is using the double angle identity for cosine.\(\displaystyle \cos2\theta=\cos^2\theta-\sin^2\theta\)

Substituting this value into the original equation gives us:

\(\displaystyle 4\cos^2\theta+\cos^2\theta-\sin^2\theta+\sin^2\theta=3\\ \implies5\cos^2\theta=3\implies\cos^2\theta=\frac{3}{5}\\ \cos\theta=\frac{\sqrt15}{5}\implies\theta=\cos^{-1}\frac{\sqrt15}{5}}\approx39.232\)

The trigonometric identity \(\displaystyle 1+cot^2(x)=csc^2(x)\), is an important identity to memorize.

Some other identities that are important to know are:

\(\displaystyle \frac{sin(x)}{cos(x)}=tan(x)\)

\(\displaystyle cot(x)=\frac{1}{tan(x)}\)

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

Factorize \(\displaystyle 0=1-cos^2\theta\).

\(\displaystyle 1-cos^2\theta= (1+cos(\theta))(1-cos(\theta))\)

Set both terms equal to zero and solve.

\(\displaystyle 1+cos(\theta)=0\)

\(\displaystyle cos(\theta)=-1\)

\(\displaystyle \theta=\pi\)

This value is not within the \(\displaystyle [0,\frac{\pi}{2}]\) domain.

\(\displaystyle 1-cos(\theta)=0\)

\(\displaystyle cos(\theta)=1\)

\(\displaystyle \theta=0\)

This is the only correct value in the \(\displaystyle [0,\frac{\pi}{2}]\) domain.

If you substitute \(\displaystyle x=sin(\theta)\) you obtain a recognizable quadratic equation which can be solved for \(\displaystyle x\), 

\(\displaystyle (x-1)^2=0\).

Then we can plug \(\displaystyle x\) back into our equation and use the unit circle to find that 

\(\displaystyle \theta=\frac{\pi}{2}\).

In order to solve \(\displaystyle 2cos^2\theta=cos(\theta)\) appropriately, do not divide \(\displaystyle cos(\theta)\) on both sides.  The effect will eliminate one of the roots of this trig function.

Substract \(\displaystyle cos(\theta)\) from both sides.

\(\displaystyle 2cos^2\theta-cos(\theta)=0\)

Factor the left side of the equation.

\(\displaystyle cos(\theta)[2cos(\theta)-1]=0\)

Set each term equal to zero, and solve for theta with the restriction \(\displaystyle [0,2\pi]\).

\(\displaystyle cos(\theta)=0\)

\(\displaystyle \theta = \frac{\pi}{2},\frac{3\pi}{2}\)

\(\displaystyle 2cos(\theta)-1=0\)

\(\displaystyle 2cos(\theta)=1\)

\(\displaystyle cos(\theta)=\frac{1}{2}\)

\(\displaystyle \theta= \frac{\pi}{3}, \frac{5\pi}{3}\)

The correct answer is:

\(\displaystyle \theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{3}, \frac{5\pi}{3}\)

By subtracting \(\displaystyle 4\) from both sides of the original equation, we get \(\displaystyle \sin^2(x) = -4\). We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.