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That part is not correct. Did you misread the given equation?2sinx+cosx + cos = 0

(2sinx)(cosx) + cosx = 0

Please try again. Hint: Factor the left-hand side.

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.

Thank for the hint.

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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

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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\(\displaystyle \text {Given } a,\ x \in \mathbb R, \ 0 < |a| \le 1,\ 0 \le x \le 2 \pi , \text { and } a * sin(x) * cos(x) + cos(x) = 0, \text { find } x.\)

Using Subhotosh's method, that general problem can be solved using the arcsin function.

\(\displaystyle a * sin(x) * cos(x) + cos(x) = 0 \implies cos(x)\{a * sin(x) + 1) = 0 \implies\)

\(\displaystyle cos(x) = 0 = \sqrt{1 - sin^2(x)} \text { or } a * sin(x) - 1 = 0 \implies sin(x) = \pm 1 \text { or } sin(x) = -\ \dfrac{1}{a}.\)

I'll bet dollars to doughnuts, however, that the specific problem is designed to be solved by special triangles. Is there any problem that

I'm going try to solve it with the arcsin function, thank!Consider the general problem:

\(\displaystyle \text {Given } a,\ x \in \mathbb R, \ 0 < |a| \le 1,\ 0 \le x \le 2 \pi , \text { and } a * sin(x) * cos(x) + cos(x) = 0, \text { find } x.\)

Using Subhotosh's method, that general problem can be solved using the arcsin function.

\(\displaystyle a * sin(x) * cos(x) + cos(x) = 0 \implies cos(x)\{a * sin(x) + 1) = 0 \implies\)

\(\displaystyle cos(x) = 0 = \sqrt{1 - sin^2(x)} \text { or } a * sin(x) - 1 = 0 \implies sin(x) = \pm 1 \text { or } sin(x) = -\ \dfrac{1}{a}.\)

I'll bet dollars to donoughts, however, that the specific problem is designed to be solved by special triangles. Is there any problem thatrequiresknowledge 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.

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Where can I buyConsider the general problem:

........I'll bet dollars todonoughts, however, that the specific problem is designed to be solved by special triangles. Is there any problem thatrequiresknowledge 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.

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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....

Please read rules of posting in this forum - enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-459460

Please share your thoughts/work with us so that we know where to begin to help you.

Please share your thoughts/work with us so that we know where to begin to help you.

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

\(\displaystyle cos(x) = 0 \implies \sqrt{cos^2(x)} = 0 \implies \sqrt{1 - sin^2(x)} = 0 \implies\)

\(\displaystyle 1 - sin^2(x) = 0 \implies sin^2(x) = 1 \implies sin(x) = \pm 1.\)

Probably you already know from the unit circle definition of the trigonometric functions a value of x such that sin(x) = 1, but suppose you do not. Then you can use the arcsin function. Set your calculator to

\(\displaystyle sin(x) = 1 \implies x = 90 \text { degrees.}\)

Now to get an exact answer in radians, use the conversion ratio of \(\displaystyle \dfrac{\pi}{180 \text { degrees}}.\)

\(\displaystyle \therefore x = 90 \text { degrees}* \dfrac{\pi}{180 \text { degrees}} = \dfrac{\pi}{2}.\)

When you do that process for sin(x) = -\ 1, you will get an answer of \(\displaystyle -\ \dfrac{\pi}{2}.\)

That answer is outside your desired range, but the sine and cosine functions are fully cyclic with a period of 2pi.

\(\displaystyle x = -\ \dfrac{\pi}{2} + 2 \pi = \dfrac{3 \pi}{2}.\)

You try it with \(\displaystyle 2sin(x) + 1 = 0.\)