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Thank you.from the question cos(x) [ cos(x) + 2sin(x) ] = 1 , so cos(x)=1

this is the question,

View attachment 13000

I need help for part (b), thanks.

As tkhunny says, there is no unit product property.

\(\displaystyle \dfrac{2}{3} * \dfrac{3}{2} = 1.\)

In this problem, it turns out that cos(x) = 1 is correct , but that is a coincidence. Moreover, cos(x) = 1 may not be the only answer.

I suspect that the restatement asked for in the problem would help, but you have not shown that you did that.

\(\displaystyle cos(x)\{cos(x) + 2 sin(x)\} = 1 \implies cos^2(x) + 2cos(x)sin(x) = 1 = cos^2(x) + sin^2(x) \implies\)

\(\displaystyle 2cos(x)sin(x) = sin^2(x).\)

\(\displaystyle sin(x) = 0 \implies x = 0 \implies cos(x) = 1.\)

And we have already confirmed that cos(x) = 1 is a valid answer. But there is another possibility, namely

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

\(\displaystyle 4cos^2(x) = 1 - cos^2(x) \implies 5cos^2(x) = 1 \implies cos^2(x) = \dfrac{1}{5} \implies\)

\(\displaystyle cos(x) = \pm \dfrac{1}{\sqrt{5}}.\)

I am not sure how you got half of this, but lets check.

\(\displaystyle cos^2(x) = \dfrac{1}{5} \implies sin^2(x) = 1 - \dfrac{1}{5} = \dfrac{4}{5} \implies sin(x) = \pm \dfrac{2}{\sqrt{5}}.\)

\(\displaystyle \pm \dfrac{1}{\sqrt{5}} * \left ( \dfrac{1}{\sqrt{5}} \pm 2 * \dfrac{2}{\sqrt{5}} \right ) = \dfrac{1}{5} + \dfrac{4}{5} = 1.\)

So \(\displaystyle cos(x) = \pm \dfrac{1}{\sqrt{5}},\ sin(x) = \pm \dfrac{2}{\sqrt{5}}\) also work.

Now what?

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Have you tried doingpart (b) has 5 answers, and I can only get some but not all 5 answers.

How to get all 5 answers byusing the expression of cos(x) + 2sin(x) = r cos (x-y)?

What do you get for r and y? If you can't do that yet, though they clearly have taught you how, please show the method they teach, as I have seen some very different approaches. It will be best if we don't confuse you with a method you haven't seen.

Once you have that, when you rewrite equation (b) using this, it will look like

r cos(x) cos(x-y) = 1

where r and y are

cos(a)cos(b) = [cos(a-b) + cos(a+b)]/2

This turns out to result in an equation you can solve directly.

Are you familiar with that formula? If not, then let us know what you

Thank you for your guidance.Have you tried doingwhat they say to do? I don't think I've seen you or anyone else do the first step. It does work, and you'll probably learn what they intend for you to learn if you try that.

What do you get for r and y? If you can't do that yet, though they clearly have taught you how, please show the method they teach, as I have seen some very different approaches. It will be best if we don't confuse you with a method you haven't seen.

Once you have that, when you rewrite equation (b) using this, it will look like

r cos(x) cos(x-y) = 1

where r and y areconstantsyou will know. What I would do next is to use the product-to-sum formula,

cos(a)cos(b) = [cos(a-b) + cos(a+b)]/2

This turns out to result in an equation you can solve directly.

Are you familiar with that formula? If not, then let us know what youhavelearned, so we can guide you along the path they have set for you.

How to solve question below?

\(\displaystyle \left|\sin2x\right|>\frac{1}{2},\left\{0<x<2\pi\right\}\)

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Is this a "new" problem?Thank you for your guidance.

How to solve question below?

\(\displaystyle \left|\sin2x\right|>\frac{1}{2},\left\{0<x<2\pi\right\}\)

If it is - please start a new thread.