5.3 – Solving Trig Equations
Dec 7th, 2010 by corricelli
Hello!
Please use this section to post questions/concerns/reflections on Chapter 5, Section 3.
Happy Blogging,
Mrs. Corricelli
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This is to mariah’s question on the chapter 5 review page:
for #s 35 and 37, use the pythagorean identities!
like for 35, change tan ^2 x to (sec^2x – 1) and solve
and for 37, after squaring the equation and subracting 1 from each side, you will have (csc^2 – 1), which is cot^2x, then you just factor and solve!
I am kind of stuck on #37 for tonight…
I’ve tried to switch csc x and cot x over into sin x and cos x terms, but it doesn’t seem to lead anywhere. Any advice?
Eample six in the chapter really helps but
cscx+cotx=1 square both sides
csc^2x + 2cscxcotx + cot^2x=1 use a pythagroean identity to substitute
cot^2x+1+2cscxcotx+cot^2x=1 Simplify
2cot^2x+2cscxcotx =0
2cotx (cotx+cscx) =0
Then you slipt them apart and solve for x. After that because you saquared in the beginning you have to check all your answers to make sure they are not extraneous. I hope this helped
I also agree with Taylor’s method, as I also used the same method, which really helped…the pythag identity workedd for both cases…but i would like to know what other way there is to solve it without the pythag. identities…or is there even another way?
im confused with 87 d and e, how do you find how many solutions the equation has and what does it mean by a greatest solution?
for #51 on the homework I graphed it and got the right graph, but when i use the calc button and try to get it to find the x intercepts it isn’t working. I put the cursor to the left of the curve and to the right of it, but when i hit enter after second curve it won’t do anything. Did anyone have this problem or know how to fix it?? thanks!
Serena, I looked at example 4 in Section 5.3 and set my window to have the maximum x-value be 2pi (make sure you’re in radians!) and then it came up with a relatively accurate graph. Then you can calculate the zeroes using [2nd trace]. (ps. know that pi/3 = 1.04)
It was in radians, and I am pretty sure about the x value being correct. Hmmm… I will try it again :p thanks Jen!