factor, solve trig eqns; find gen. solns; evaluate; etc

[Tony]

I have a trig test tomorrow and I'm absolutely clueless as to how to solve these problems. ANY help at all would be greatly appreciated.

My teacher gave us some sample problems and I have no idea how to do them. Here are a few: (pi is pie 3.14)

1) Factor and find all solutions of the equation 2cot^2x - 3 csc x = 0 on the interval [0,2pi)

I'm unsure whether I add 3 csc x to the opposite side or not. That's pretty much how far I got with that.

2) Find all solutions of the equation 2sin^2 3x - 1 = 0 on the interval [0,pi).

Hint: there are 6 of these. Hint: Find specific solutions on [0,2pi). Write general solutions.

I'm sorry, I haven't been able to even start this problem.

3) Find all solutinos of the equation: 2 cosxsinx = sinx

I did the following:

sinx (2 cosx - 1) = 0
sinx = 0
cosx = 1/2
pi/3 + n(2pi)
5pi/3 + n(2pi)
x = 0, pi, 2pi

I don't know if that's right or not.

4) Find the exact value of the expression cos (pi/12) cos(5pi/12) + sin (pi/12) sin (5pi/12)

I have no idea.

5) Find the exact solutions to the equation 2sin2x = 2cosx on the interval [0,2pi)

Once again, I have no idea.

6) Find the exact values of the sine, cosine, and tangent of the angle 11pi/12 , given that 11pi/12 = 3pi/4 + pi/6

Do you replace 3pi/4 with whatever the cosine, sine, and tangent variable is (-root 2/2, root 2/2, and -1), and do the same for pi/6?

7) Given that sin u= 15/17 and cos v = -40/41 and both angles u and v are in Quadrant II. Find the exact value of each expression. Hint: Use right triangles.

a) sin(u + v)
b) cos(u + v)

I got as far as writing the right triangles.

That's about it. I really appreciate what you guys do here and any help on these problems is very welcome. Thank you.

 

[soroban]

Re: URGENT TRIG HELP

Hello, Tony!

Here are a few of them.
. . Good luck on your exam.

Factor and find all solutions of the equation on the interval \(\displaystyle [0,\,2\pi)\)
. . \(\displaystyle 1)\;2\cdot\cot^2x\,-\,3\cdot\csc x\:=\:0\)

We're expected to know the identity: \(\displaystyle \,\cot^2x\:=\:\csc^2x\,-\,1\)

The equation becomes: \(\displaystyle \:2(\csc^2x\,-\,1)\,-\,3\cdotcsc x \;=\;0\)

. . and we have a quadratic: \(\displaystyle \:2\cdot\csc^2x\,-\,3\cdot\csc x \,-\,2\;=\;0\)

. . which factors: \(\displaystyle \2\cdot\csc x\,-\,1)(\csc x \,-\,2)\;=\;0\)


And we have two equations to solve:

. . \(\displaystyle 2\cdot\csc x\,-\,1\:=\:0\;\;\Rightarrow\;\;\csc x\:=\:\frac{1}{2}\) . . . no real solutions

. . \(\displaystyle \csc x \,-\,2\:=\:0\;\;\Rightarrow\;\;\csc x\:=\:2\;\;\Rightarrow\;\;\L\fbox{x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}}\)

Find all solutions of the equation on the interval \(\displaystyle [0,\,\pi)\).
. . \(\displaystyle 2)\;2\cdot\sin^23x\,-\,1\:=\:0\)

We have: \(\displaystyle \:\sin^23x\:=\:\frac{1}{2}\;\;\Rightarrow\;\;\sin3x\:=\:\pm\frac{1}{\sqrt{2}}\)

Then: \(\displaystyle \:3x\;=\;\frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{4},\:\frac{7\pi}{4},\:\frac{9\pi}{4},\:\frac{11\pi}{4}\)

Therefore: \(\displaystyle \L\:\fbox{x\;=\;\frac{\pi}{12},\;\frac{\pi}{4},\:\frac{5\pi}{12},\:\frac{7\pi}{12},\:\frac{3\pi}{4},\:\frac{11\pi}{12}}\)

Find all solutions of the equation:
. . \(\displaystyle 3)\;2\cdot\cos x\cdot\sin x\:=\:\sin x\)

Your work was pretty good . . . your answers could be neater.

We have: \(\displaystyle \:2\cdot\cos x\cdot\sin x \,-\,\sin x\:=\:0\)

Factor: \(\displaystyle \:\sin x(2\cdot\cos x\,-\,1)\:=\:0\)

Solve: \(\displaystyle \:sin x\:=\:0\;\;\Rightarrow\;\;\L\fbox{x\:=\:\pi n}\)

Solve: \(\displaystyle \:2\cdot\cos x\,-\,1\:=\:0\;\;\Rightarrow\;\;\cos x\:=\:\frac{1}{2}\;\;\Rightarrow\;\;\L\fbox{x \:=\:\pm\frac{\pi}{3}\,+\,2\pi n}\)

Find the exact value of the expression:
. . \(\displaystyle 4)\;\cos(\frac{\pi}{12})\cdot\cos(\frac{5\pi}{12})\,+\,\sin(\frac{\pi}{12})\cdot\sin(\frac{5\pi}{12})\)

We're expected to recognize this compound angle identity:

. . \(\displaystyle \cos(A\,-\,B) \:=\:\cos(A)\cdot\cos(B)\,+\,\sin(A)\cdot\sin(B)\)


So that: \(\displaystyle \:\cos\left(\frac{\pi}{12}\right)\cdot\cos\left(\frac{5\pi}{12}\right)\,+\,\sin\left(\frac{\pi}{12}\right)\cdot\sin\left(\frac{5\pi}{12}\right) \;=\;\cos\left(\frac{\pi}{12}\,-\,\frac{5\pi}{12}\right)\)

. . \(\displaystyle = \;\cos\left(-\frac{4\pi}{12}\right) \;=\;\cos\left(-\frac{\pi}{3}\right) \;=\;\cos\left(\frac{\pi}{3}\right) \;=\;\L\fbox{\frac{1}{2}}\)