There are many ways to solve this problem, and here is one.

First, re-write the equation so that it is set equal to zero, since we're finding the roots, and so that it is all in one fraction. That is easy to do since both terms already have the common denominator of 3:

\(\displaystyle \frac{ \sqrt3 - 2 \sin (\frac{\theta }{2} ) } {3 } = 0\) multiply both sides by 3

\(\displaystyle \sqrt3 - 2 \sin (\frac{ \theta }{ 2 }) = 0\) add the sine term to both sides

\(\displaystyle \sqrt3 = 2 \sin (\frac{ \theta }{2 })\) divide both sides by 2

\(\displaystyle \frac{ \sqrt3 }{2} = \sin (\frac{ \theta }{2} )\)

\(\displaystyle \sin ^ {-1 } (\frac{\sqrt3}{2} ) = \frac{ \theta }{2}\)

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

\(\displaystyle \theta = \frac{ 2 \pi }{3} , \frac{ 4 \pi }{3}\)

Factor the equation by grouping: the first 2 terms have \(\displaystyle \sin \theta\) in common, and the second 2 have 5 in common:

\(\displaystyle y = \sin \theta [ 2 \cos ^2 (\frac{ 2}{3 } \theta ) - 1 ] + 5 [ 2 \cos ^2 (\frac{2}{3} \theta ) - 1 ]\)

\(\displaystyle y = (2 \cos ^2 (\frac{ 2}{3} \theta ) - 1 ) ( \sin \theta + 5 )\)

Since we are finding the roots, we can set each factor equal to 0:

\(\displaystyle \sin \theta + 5 = 0\) subtract 5 from both sides

\(\displaystyle \sin \theta = -5\) this is outside of the range for sine

\(\displaystyle 2 \cos ^2 (\frac{ 2}{3} \theta ) - 1 = 0\) add 1 to both sides

\(\displaystyle 2 \cos ^ 2 ( \frac{ 2}{3 } \theta ) = 1\)divide both sides by 2

\(\displaystyle \cos ^2 ( \frac{ 2}{3 }\theta ) = \frac{ 1}{2}\) take the square root of both sides

\(\displaystyle \cos (\frac{ 2}{3} \theta ) = \pm \frac{ 1 }{\sqrt2}\)

\(\displaystyle \frac{2}{3} \theta = \cos ^ {-1} ( \pm \frac{ 1}{ \sqrt2})\)

\(\displaystyle \frac{ 2}{3} \theta = \frac{ \pi }{4} , \frac{ 7 \pi }{4} , \frac{ 3 \pi }{4} , \frac{ 5 \pi }{4}\) multiply by \(\displaystyle \frac{ 3}{2 }\)

\(\displaystyle \theta = \frac{ 3 \pi }{8 } , \frac{ 21 \pi }{8 } , \frac{ 9 \pi }{8 } , \frac{ 15 \pi }{8 }\)

That second one is greater than \(\displaystyle 2\pi\), so we want to figure out which angle between 0 and \(\displaystyle 2 \pi\) is coterminal with it. We can figure this out by subtracting \(\displaystyle 2 \pi\), or in this case \(\displaystyle \frac{ 16 \pi }{ 8 }\) to get \(\displaystyle \frac{ 21 \pi }{ 8 } - \frac{ 16 \pi }{ 8 } = \frac{ 5 \pi } {8 }\)

Factor by grouping. The first 2 terms have \(\displaystyle \cos(4 \theta)\) in common.

\(\displaystyle y = \cos (4\theta ) [ 2 \sin \theta - \sqrt3 ] + 1[2 \sin \theta - \sqrt3]\)

\(\displaystyle y = ( \cos (4 \theta ) + 1 ) ( 2 \sin \theta - \sqrt3 )\)

We are finding the roots, so set each factor equal to 0:

\(\displaystyle \cos (4 \theta ) + 1 = 0\) subtract 1 from both sides

\(\displaystyle \cos (4 \theta ) = -1\)

\(\displaystyle 4 \theta = \cos ^ {-1 } (-1)\)

\(\displaystyle 4 \theta = \pi\) divide by 4 

\(\displaystyle \theta = \frac{ \pi }{4}\)

\(\displaystyle 2 \sin \theta - \sqrt3 = 0\) add \(\displaystyle \sqrt3\) to both sides

\(\displaystyle 2 \sin \theta = \sqrt 3\) divide both sides by 2

\(\displaystyle \sin \theta = \frac{ \sqrt3}{2 }\)

\(\displaystyle \theta = \sin ^ { -1 } (\frac{ \sqrt3}{2} )\)

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

Factor the equation:

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

The roots occur when each factor equals 0:

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

\(\displaystyle \sin x = 1\)

\(\displaystyle x = \sin ^{-1}(1) = \frac{\pi }{2}\)

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

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

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

\(\displaystyle x = \cos ^{-1}(\frac{1}{2}) = \frac{\pi }{6} , \frac{5 \pi }{6}\)

Take the square root of both sides:

\(\displaystyle \tan^2 x=3/4\rightarrow\tan x=\pm\sqrt{3}/2\)

This gives us two roots, 

\(\displaystyle \tan x=\sqrt{3}/2\) and \(\displaystyle \tan x=-\sqrt{3}/2\)

Solving for x:

\(\displaystyle x=\arctan (\sqrt{3}/2)\) or \(\displaystyle x=\arctan (-\sqrt{3}/2)\),

which evaluate to \(\displaystyle x=0.71\) and \(\displaystyle x=-0.71\)

Now we check to make sure both answers work. Make sure your calculator is in radians mode!

\(\displaystyle \tan^2 (0.71)=0.74\approx 3/4\)

\(\displaystyle \tan^2 (-0.71)=0.74\approx 3/4\)

Our answers are \(\displaystyle x=0.71, x=-0.71\)

if \(\displaystyle \small \tan x\) is \(\displaystyle \small \frac{\sin x }{\cos x }\) 

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

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

\(\displaystyle x=\sin^{-1}(1)=90\)

\(\displaystyle 3\sin x=\cos 2x\);  Use the double angle identity for cosine.

\(\displaystyle 3\sin x=1-2\sin ^2x\);  Move everything to the left side of the equation.

\(\displaystyle 2\sin ^2x+3\sin x-1=0\);  This is a quadratic-like expression that cannot be factored.  We must use the quadratic formula.  It may be helpful to see this if you replace \(\displaystyle \sin x\) with \(\displaystyle y\), so it becomes:

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

Recall the quadratic formula  \(\displaystyle y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

plug in \(\displaystyle a=2; b=3; c=1\).

We now have

\(\displaystyle \sin x = \frac{-3\pm \sqrt{17}}{4}\);  Separate this into two equations and take the inverse sine.

\(\displaystyle x = \sin ^{-1}\left( \frac{-3+\sqrt{17}}{4}\right )\) or  \(\displaystyle x = \sin ^{-1}\left( \frac{-3-\sqrt{17}}{4}\right )\)

The first equation gives us \(\displaystyle x=0.285\).  Using the unit circle as we did in previous problems, we can find a second answer from this which is \(\displaystyle x=2.857\).  The second equation will not give us a solution.

When the quadratic formula is applied to the function, it yields

\(\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-5\pm \sqrt{25-16}}{4}=\frac{5\pm3}{4}= =-\frac{1}{2}, -2\)

So those are the zeros for sine, but sine has a minimum of -1, so -2 is out. For -1/2, sine achieves that twice in a cycle, at π+π/6 and 2π-π/6. So while -π/6 is true, it is not correct since it is not in the given interval.

Therefore on the given interval the zeros are:

\(\displaystyle \frac{7\pi}{6}, \frac{11\pi}{6}\)

Since \(\displaystyle cos^{2}(x)+cos(x)-2=0\) can be written as:

\(\displaystyle (cosx-1)(cosx+2)=0\). We can't have \(\displaystyle cosx=-2\).

Therefore \(\displaystyle cosx-1=0\) . This means that \(\displaystyle x=2k\pi,\)where k is an integer.

since \(\displaystyle \in[0,\p\frac{\pi}{2}]\). We have x=0 is the only number that satisfies this property.

Rearrange the problem,

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

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

Over the interval 0 to 360 degrees, cosx = 1/2 at 60 degrees and 300 degrees.