Compute Trig Function Values, Solve Trig Equation

I need a little help here !!!

Find the following values exactly:

a) sin (7pi/6) I got -1/2 but i don't know how i got it please someone explain

b) sin (-11pi/4)

2) Find the general soln of the trig eqn

3cosx = 2sin^2x

My answer is : (2 cos x - 1) (cos x + 2) is that right ?

Re: Compute Trig Function Values, Solve Trig Equation

Firstly, $\displaystyle \frac{7\pi}{6}$ is equivalent to 210 degrees (multiply by 180/pi). 210 degrees has a "reference angle" of 30, meaning that it has the same sin/cos/tan as 30 degrees except that it is in a different quadrant and thus may be negative. In this case, the sine of 210 degrees is **negative**, because sine is negative in the third quadrant.

This means that $\displaystyle \sin \frac{7\pi}{6}=\sin 210^\circ = -\sin 30^\circ = -\frac{1}{2}$

See if you can do part b) yourself by using the same strategy--convert to degrees if it makes things easier for you, find a reference angle, and determine whether sine should be positive or negative in the appropriate quadrant.

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For part 2), I think you misunderstand what the problem is asking. The problem is asking for the numbers (angles) such that the equation is true. So trig functions should not appear in the answer. Hint: rewrite $\displaystyle \sin^2 x$ as $\displaystyle 1-\cos^2 x$, bring all the terms to one side, factor, and solve.