\(\displaystyle 3\sin x=\cos 2x\);  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. 

First, get the equation in terms of one trig function. We can do this by substituting in using the Pythagorean Identity for \(\displaystyle cos^2x\).

Then we have \(\displaystyle 1-sin^2x = 2sinx +2\).

Bring all the terms to one side to find that \(\displaystyle sin^2x+2sinx+1\).

We can factor this quadratic to \(\displaystyle (sinx+1)^2 = 0\).

This means that \(\displaystyle sinx = -1\).

The only angle value for which this is true is \(\displaystyle \frac{3\pi}{2}\). 

First, re-write the equation so that it is equal to zero: \(\displaystyle sin^2x + 2sinx -3 =0\)

Now we can use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=1, b=2, and c=-3:

\(\displaystyle sinx = \frac{-2\pm\sqrt{(-2)^2-(4\cdot 1\cdot -3)}}{2\cdot 1}\)simplify

\(\displaystyle sinx = \frac{-2\pm\sqrt{4+12}}{2}\)

\(\displaystyle sinx = \frac{-2\pm\sqrt{16}}{2}\)

\(\displaystyle sinx=\frac{-2\pm4}{2}\)

This gives two potential answers:

\(\displaystyle sinx=\frac{-2+4}{2}=\frac{2}{2}=1\)

and

\(\displaystyle sinx=\frac{-2-4}{2}=\frac{-6}{2}=-3\)

Sine must be between -1 and 1, so there are no values of x that would give a sine of -3. The only solution that works is \(\displaystyle sinx=1\). The only angle measure that has a sine of 1 is \(\displaystyle 90^o\).

Use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=4, b=1, and c=-1:

\(\displaystyle sinx = \frac{-1\pm\sqrt{(-1)^2-(4\cdot 4\cdot -1)}}{2\cdot 4}\)simplify

\(\displaystyle sinx = \frac{-1\pm\sqrt{1+16}}{8}\)

\(\displaystyle sinx = \frac{-1\pm\sqrt{17}}{8}\)

the square root of 17 is about 4.123. This gives two potential answers:

\(\displaystyle sinx=-0.64, or 0.39\). We can solve for x by evaluating both \(\displaystyle sin^{-1}(-0.64)\) and \(\displaystyle sin^{-1}(0.39)\). The first gives an answer of \(\displaystyle -39.79^o\). Add this to 360 to get that as a positive angle measure, \(\displaystyle 320.21^o\). If this has a sine of -0.64, so does its reflection over the y-axis, which is \(\displaystyle 219.79^o\).

The second gives an answer of \(\displaystyle 22.95^o\). If that has a sine of 0.39, then so does its reflection over the y-axis, which is \(\displaystyle 157.05^o\).

There are multiple solution paths. We could subtract 1 from both sides and use the quadratic formula with \(\displaystyle a = 2, b = 0,\) and \(\displaystyle c = -1\). Or we could solve using inverse opperations:

\(\displaystyle 2 \sin ^2 (3 \theta) = 1\) divide both sides by 2

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

\(\displaystyle \sin (3 \theta ) = \pm \frac{1} { \sqrt 2 }\) 

The unit circle tells us that potential solutions for \(\displaystyle 3 \theta\) are \(\displaystyle \frac{\pi }{4}, \frac{3 \pi }{4} , \frac{5 \pi }{4} , \frac{7 \pi }{4}\).

To get our final solution set, divide each by 3, giving:

\(\displaystyle \frac{ \pi }{12}, \frac{\pi }{4 }, \frac{5 \pi }{12}, \frac{ 7 \pi }{12 }\).

 This problem has multiple solution paths, including subtracting 5 from both sides and using the quadratic formula with \(\displaystyle a = 4, b = 0 , c = -3\). We can also solve using inverse opperations:

\(\displaystyle 4 \sin ^2 (\theta + \frac{ \pi }{3} ) + 2 = 5\) subtract 2 from both sides

\(\displaystyle 4 \sin ^2 (\theta + \frac{ \pi }{3} ) = 3\) divide both sides by 4

\(\displaystyle \sin ^ 2 (\theta + \frac{ \pi }{3} ) = \frac{3}{4}\) take the square root of both sides

\(\displaystyle \sin (\theta + \frac{ \pi }{3} ) = \pm \frac{\sqrt3}{2}\)

If the sine of an angle is \(\displaystyle \pm \frac{\sqrt 3 }{2}\), that angle must be one of \(\displaystyle \frac{ \pi }{3} , \frac{2 \pi }{3} , \frac{4 \pi }{3} , \frac{5 \pi }{3}\). Since the angle is \(\displaystyle \theta + \frac{ \pi }{3}\), we can get theta by subtracting \(\displaystyle \frac{ \pi }{3}\):

\(\displaystyle \frac{\pi }{3} - \frac{ \pi }{3} = 0\)

\(\displaystyle \frac { 2 \pi }{3} - \frac{ \pi }{3} = \frac{ \pi }{3}\)

\(\displaystyle \frac{ 4 \pi }{3} - \frac{ \pi }{3} = \frac{ 3 \pi }{3} = \pi\)

\(\displaystyle \frac{ 5 \pi }{3} - \frac{ \pi }{3} = \frac{ 4 \pi }{3}\)

To solve, use the quadratic formula with \(\displaystyle a = 4, b = -4, c = -3\) and \(\displaystyle \cos \theta\) where x would normally be:

\(\displaystyle \cos \theta = \frac{4 \pm \sqrt{16 -4(4)(-3)}}{2(4)} = \frac{4 \pm 8}{8 }\)

This gives us two potential answers:

\(\displaystyle \cos \theta = \frac{4 + 8 }{8 } = \frac{12}{8 } = 1.5\) since this number is greater than 1, it is outside of the domain for cosine and won't give us any solutions.

\(\displaystyle \cos \theta = \frac{4 - 8 }{8} = \frac{ -4} { 8 } = \frac{-1}{2}\)

Consulting the unit circle, the cosine is \(\displaystyle -\frac{1}{2}\) when \(\displaystyle \theta = \frac{2 \pi }{3} , \frac{ 4 \pi }{3}\)