# Unable to get correct answer

#### mimie

##### New member
Please teach me how to get correct answer. Thanks.

#### JeffM

##### Elite Member
It would really help to know the question.

And how did you conclude that the answer required cos(x) to be 1. That leads to one set of answers, but does it lead to every answer?

#### mimie

##### New member
It would really help to know the question.

And how did you conclude that the answer required cos(x) to be 1. That leads to one set of answers, but does it lead to every answer?
from the question cos(x) [ cos(x) + 2sin(x) ] = 1 , so cos(x)=1

this is the question,

I need help for part (b), thanks.

#### tkhunny

##### Moderator
Staff member
That multiplication/factoring thing is a property of zero (0), not of one (1).

If A*B = 0, then either A = 0 or B = 0.

If A*B = 1, there is no such thing.

#### JeffM

##### Elite Member
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.
Thank you.

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?

#### mimie

##### New member
from 2cos(x)sin(x) = sin^2(x), how to get sin(x)=0 ?
part (b) has 5 answers, and I can only get some but not all 5 answers.
How to get all 5 answers by using the expression of cos(x) + 2sin(x) = r cos (x-y) ?

#### Dr.Peterson

##### Elite Member
part (b) has 5 answers, and I can only get some but not all 5 answers.
How to get all 5 answers by using the expression of cos(x) + 2sin(x) = r cos (x-y) ?
Have you tried doing what 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 are constants you 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 you have learned, so we can guide you along the path they have set for you.

#### mimie

##### New member
Have you tried doing what 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 are constants you 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 you have learned, so we can guide you along the path they have set for you.
Thank you for your guidance.

How to solve question below?
$$\displaystyle \left|\sin2x\right|>\frac{1}{2},\left\{0<x<2\pi\right\}$$

Last edited:

#### Subhotosh Khan

##### Super Moderator
Staff member
Thank you for your guidance.

How to solve question below?
$$\displaystyle \left|\sin2x\right|>\frac{1}{2},\left\{0<x<2\pi\right\}$$
Is this a "new" problem?

If it is - please start a new thread.