1. ## Trigonometric equations

Solve for x
1. $\displaystyle sin3x= \frac{3}{4}$
2. $\displaystyle 4sinx.cosx.cos2x=1$
3. $\displaystyle -cos4x=2$
4. $\displaystyle cos(x+\frac{pi}{18})=\frac{2}{5}$

2. For the second one, you need to know one identity in particular.

$\displaystyle sin(2A) = 2sinAcosA$

[4sin(x)cos(x)]cos(2x) = [2sin(2x)]cos(2x) = sin(4x)

Then, you solve for x.

For that, you find the first value of x as being the inverse of sine.

I'll call this angle alpha.

$\displaystyle \alpha = sin^{-1}(1) = 90^o$

Then, you need to know in which quadrant sin is positive. It is positive in the first and second quadrant.

So, the value of 4x becomes 90 degrees.

Now, since x has a coefficient of 4, you need to consider 4 cycles, that is, you need to include 90+360, 90+720 and 90+1080.

This gives:

4x = 90, 450, 810, 1170 degrees

So, now you can find the value of x for each of those values. Note that I'm taking x to be within the interval of $\displaystyle 0^o \leq x \leq 360^o$