That part is not correct. Did you misread the given equation?2sinx + cosx + cos = 0
yes you right, i meant to type 2sinx + cosx + cosx = 0That's not correct. Did you misread the given equation?
(2sinx)(cosx) + cosx = 0
Please try again. Hint: Factor the left-hand side.
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You changed 'cos' to 'cosx' above (fixing a typo that I had ignored), but that plus sign is still wrong, heh.… i meant to type 2sinx + cosx + cosx = 0 …
omg you have some good eyes lol, correction 2sinx cosx + cosx = 0You changed 'cos' to 'cosx' above (fixing a typo that I had ignored), but that plus sign is still wrong, heh.
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Now factor out cos(x) to get:omg you have some good eyes lol, correction 2sinx cosx + cosx = 0
I'm going try to solve it with the arcsin function, thank!Consider the general problem:
[MATH]\text {Given } a,\ x \in \mathbb R, \ 0 < |a| \le 1,\ 0 \le x \le 2 \pi , \text { and } a * sin(x) * cos(x) + cos(x) = 0, \text { find } x.[/MATH]
Using Subhotosh's method, that general problem can be solved using the arcsin function.
[MATH]a * sin(x) * cos(x) + cos(x) = 0 \implies cos(x)\{a * sin(x) + 1) = 0 \implies[/MATH]
[MATH]cos(x) = 0 = \sqrt{1 - sin^2(x)} \text { or } a * sin(x) - 1 = 0 \implies sin(x) = \pm 1 \text { or } sin(x) = -\ \dfrac{1}{a}.[/MATH]
I'll bet dollars to donoughts, however, that the specific problem is designed to be solved by special triangles. Is there any problem that requires knowledge of special triangles for solution? I don't get the pedagogic value of teaching a method that is far from general when a general method exists and takes only a decent calculator to apply.
Where can I buy those?!?Consider the general problem:
........I'll bet dollars to donoughts, however, that the specific problem is designed to be solved by special triangles. Is there any problem that requires knowledge of special triangles for solution? I don't get the pedagogic value of teaching a method that is far from general when a general method exists and takes only a decent calculator to apply.
What answers did you get by following the hint above?Now factor out cos(x) to get:
2sin(x)* cos(x) + cos(x) = 0
cos(x) * [2sin(x) + 1] = 0 Thus
Either
cos(x) = 0
or
2sin(x) + 1 = 0
Continue....
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OK. You followed Subhotosh's explanation of the factoring and the use of the zero product property, right? You come up with two possibilities. Let's follow through on one of them. I am going to assume that you are working in radians.no luck on this problem yet! smh