(tan^2x-9)(2cosx+1)=0 Find one period
4cos^2x-4cosx+1=0 find all angles and name 8 angles
7sin^2x-22sinx+3=0 find all angles and name 8 angles
2cos^2x-7cosx=-3 find one period
sinx(2sinx+1)=0 find one period
I already took this course. Please try the problems before asking us to do them all.
if i had any idea about these problems i wouldnt ask you in the first place.
@Adam: if these kinds of problems are hard for you to understand (i know they are for me)try algebra-dot-com. Type your question in the google custom search box. They have some problems in the archives. If they don't have the ones you're looking for, go back to the homepage, pick Algebra II: Trig from the topic lists on the right, and click ask free tutor. You'll get an answer in a few hours.
Adam: I suggest you post one problem, see how it is solved. Try the others along the same line and ask another if necessary. There is a learning process in doing these problems. You are taking the course, not Damon or anyone else.
Now some hints for you:
1. First try to transform the ratios to sin or cos, and next into only sin or only cos.
Use identities to do that, for example,
cos²x+sin²x=1
1+tan²x=1/cos²x
sin(2x)=2sinx cosx
cos(2x)=cos²x-sin²x
and so on. This is part of the learning process to be aware of these available tools.
2. If you are able to transform all the ratios to, say cosx and its powers, then do the substitution c=cos(x) and solve the polynomial in x.
3. If you can factorize the expression on the left, solve each factor separately. Make a union of the solutions as the final solution.
More hints:
(tan^2x-9)(2cosx+1)=0 Find one period
- factorized already, solve separately
4cos^2x-4cosx+1=0 find all angles and name 8 angles
- expand cos²x and use the identity sin²+cos²=1 to convert sin to cos.
7sin^2x-22sinx+3=0 find all angles and name 8 angles
-use the double angle formula
2cos^2x-7cosx=-3 find one period
-this is just a polynomial in cos(x). Substitute c=cos(x) and solve for c.
sinx(2sinx+1)=0 find one period
This is already factored. Solve each factor and union the solutions.
If you still have problems, post your attempt(s).
Sorry, I have misread the questions confusing cos^2x for cos2x.
Actually, most of the problems are simply polynomials in sin(x) or cos(x).
After the first problem, make the substitutions s=sin(x) or c=cos(x) and solve the resulting polynomial (quadratic) in s or c.
i'm so confused right now. can u solve one question for me so i can try the rest?
thank you!
OK.
4cos^2x-4cosx+1=0 find all angles and name 8 angles
Let c=cos(x), then the equation becomes
4c²-4c+1=0
(2c-1)²=0
So c=1/2, or cos(x)=1/2
If you look at the graph of cos(x), you will find that cos(x)=1/2 occurs at π/3 and 5π/3. So the general solution is
x=π/3+2kπ or 5π/3+2kπ, where k is an integer.