# Can I divide both sides by tan(x)?

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Hello

I have problem from the A-level revision book, and I have manipulated the problem until I reach this point (everything is correct up to here):

tan(x) = √2 × sin(x) tan (x)

I'm thinking that I can divide both sides by tan(x), which will give me:

1 = √2 × sin(x)

The answer section of the book arrives at the same point, but via a different manipulation of the trig identity: tan(x) = sin(x) / cos(x). But the book seems to divide both sides by cos(x), leaving (like me): √2 = 1 / sin(x). I used the same identity, but I manipulated it differently, originally. I have arrived at the same answer.

My question is: can/should I divide both sides of an equation like this by a trig function of an unknown? Because I know that by dividing both sides of an equation by an unknown will remove a possible answer. For example, in other problems, it could remove the possible answer of x=0. Is that the case here? Am I losing possible answers?

Thanks in advance

I have problem from the A-level revision book, and I have manipulated the problem until I reach this point (everything is correct up to here):

tan(x) = √2 × sin(x) tan (x)

I'm thinking that I can divide both sides by tan(x), which will give me:

1 = √2 × sin(x)

The answer section of the book arrives at the same point, but via a different manipulation of the trig identity: tan(x) = sin(x) / cos(x). But the book seems to divide both sides by cos(x), leaving (like me): √2 = 1 / sin(x). I used the same identity, but I manipulated it differently, originally. I have arrived at the same answer.

My question is: can/should I divide both sides of an equation like this by a trig function of an unknown? Because I know that by dividing both sides of an equation by an unknown will remove a possible answer. For example, in other problems, it could remove the possible answer of x=0. Is that the case here? Am I losing possible answers?

Thanks in advance

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#2

(Original post by

...

**the81kid**)...

so that you have two solution families: which gets you one set and which gets you another.

You should not divide by , that will definitely lose you solutions.

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(Original post by

You should see and your immediate reaction should be:

so that you have two solution families: which gets you one set and which gets you another.

You should not divide by , that will definitely lose you solutions.

**Zacken**)You should see and your immediate reaction should be:

so that you have two solution families: which gets you one set and which gets you another.

You should not divide by , that will definitely lose you solutions.

Yes, that was my first thought too, that i'm losing solutions. I was going to make it a quadratic equation, then I can keep both solutions. But the book has only gone with the eventual solution of: π/4=x. I don't know why.

PS. How do I type equations in these questions/comments? I can't find the button or option when I'm writing a question.

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#4

(Original post by

PS. How do I type equations in these questions/comments? I can't find the button or option when I'm writing a question.

**the81kid**)PS. How do I type equations in these questions/comments? I can't find the button or option when I'm writing a question.

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#5

(Original post by

Many thanks again for the quick reply!

Yes, that was my first thought too, that i'm losing solutions. I was going to make it a quadratic equation, then I can keep both solutions. But the book has only gone with the eventual solution of: π/4=x. I don't know why.

**the81kid**)Many thanks again for the quick reply!

Yes, that was my first thought too, that i'm losing solutions. I was going to make it a quadratic equation, then I can keep both solutions. But the book has only gone with the eventual solution of: π/4=x. I don't know why.

I don't see how this could be a quadratic equation, though. It's just factorising and setting each factor equal to zero in turn.

PS. How do I type equations in these questions/comments? I can't find the button or option when I'm writing a question.

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(Original post by

No problem! The book seems quite awry - to be honest. You'd get more than one solution (depending on the range).

I don't see how this could be a quadratic equation, though. It's just factorising and setting each factor equal to zero in turn.

Don't worry about it, you need to learn some typesetting code called but nobody minds if you continue to typeset equations using ASCII (as in the form sqrt(2) = sin x, it's perfectly readable).

**Zacken**)No problem! The book seems quite awry - to be honest. You'd get more than one solution (depending on the range).

I don't see how this could be a quadratic equation, though. It's just factorising and setting each factor equal to zero in turn.

Don't worry about it, you need to learn some typesetting code called but nobody minds if you continue to typeset equations using ASCII (as in the form sqrt(2) = sin x, it's perfectly readable).

I can more or less type LaTeX (just have to remember the tex tags).

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#7

(Original post by

I'm a little worried about the book. It's the CGP revision book, and it's highly recommended. Ah yes, it's not a quadratic, I misspoke. Just 0 = 2 factors. I solved it as you said, and I also get the answer x=0. The question says: solve the following equation for -π≤x≤π. Correct me if I'm missing a glaring mistake, but x=0 satisfies that answer, no?

**the81kid**)I'm a little worried about the book. It's the CGP revision book, and it's highly recommended. Ah yes, it's not a quadratic, I misspoke. Just 0 = 2 factors. I solved it as you said, and I also get the answer x=0. The question says: solve the following equation for -π≤x≤π. Correct me if I'm missing a glaring mistake, but x=0 satisfies that answer, no?

I can more or less type LaTeX (just have to remember the tex tags).

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(Original post by

Yep, is a solution, don't forget that are also solutions. Like I said yesterday, the sine question was a mistake on the textbook authors part and so is this, seemingly.

Ah, then all you need to do is type: [tex]\LaTeX code[/tex].

**Zacken**)Yep, is a solution, don't forget that are also solutions. Like I said yesterday, the sine question was a mistake on the textbook authors part and so is this, seemingly.

Ah, then all you need to do is type: [tex]\LaTeX code[/tex].

I drew a sine graph sketch, and the only answers are positive (except the x=0 of course, which comes from tan(x)=0).

(I'll try my hand at this LaTeX thing next question.)

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#9

(Original post by

Hello

I have problem from the A-level revision book, and I have manipulated the problem until I reach this point (everything is correct up to here):

tan(x) = √2 × sin(x) tan (x)

**the81kid**)Hello

I have problem from the A-level revision book, and I have manipulated the problem until I reach this point (everything is correct up to here):

tan(x) = √2 × sin(x) tan (x)

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#10

(Original post by

Hello

I have problem from the A-level revision book, and I have manipulated the problem until I reach this point (everything is correct up to here):

tan(x) = √2 × sin(x) tan (x)

I'm thinking that I can divide both sides by tan(x), which will give me:

1 = √2 × sin(x)

The answer section of the book arrives at the same point, but via a different manipulation of the trig identity: tan(x) = sin(x) / cos(x). But the book seems to divide both sides by cos(x), leaving (like me): √2 = 1 / sin(x). I used the same identity, but I manipulated it differently, originally. I have arrived at the same answer.

My question is: can/should I divide both sides of an equation like this by a trig function of an unknown? Because I know that by dividing both sides of an equation by an unknown will remove a possible answer. For example, in other problems, it could remove the possible answer of x=0. Is that the case here? Am I losing possible answers?

Thanks in advance

**the81kid**)Hello

I have problem from the A-level revision book, and I have manipulated the problem until I reach this point (everything is correct up to here):

tan(x) = √2 × sin(x) tan (x)

I'm thinking that I can divide both sides by tan(x), which will give me:

1 = √2 × sin(x)

The answer section of the book arrives at the same point, but via a different manipulation of the trig identity: tan(x) = sin(x) / cos(x). But the book seems to divide both sides by cos(x), leaving (like me): √2 = 1 / sin(x). I used the same identity, but I manipulated it differently, originally. I have arrived at the same answer.

My question is: can/should I divide both sides of an equation like this by a trig function of an unknown? Because I know that by dividing both sides of an equation by an unknown will remove a possible answer. For example, in other problems, it could remove the possible answer of x=0. Is that the case here? Am I losing possible answers?

Thanks in advance

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#11

(Original post by

Since x=0 is possible, will they want this answer in the exam too?

**the81kid**)Since x=0 is possible, will they want this answer in the exam too?

I drew a sine graph sketch, and the only answers are positive (except the x=0 of course, which comes from tan(x)=0).

So all in all, your solutions should be .

You can plug these into your original equation to see that they're all correct.

(I'll try my hand at this LaTeX thing next question.)

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(Original post by

Definitely!

.

So all in all, your solutions should be .

You can plug these into your original equation to see that they're all correct.

Awesome!

**Zacken**)Definitely!

.

So all in all, your solutions should be .

You can plug these into your original equation to see that they're all correct.

Awesome!

Thanks again, I'll remember all this for the exam.

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#13

(Original post by

x also equals with the symmetry thing going on.

Thanks again, I'll remember all this for the exam.

**the81kid**)x also equals with the symmetry thing going on.

Thanks again, I'll remember all this for the exam.

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(Original post by

What is the original question? You may have introduced additional solutions by your working.

**ghostwalker**)What is the original question? You may have introduced additional solutions by your working.

The original question is: √2 × cos(x) = 1 / tan(x)

The book seems to have missed a possible answer.

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(Original post by

Yep. Just plug them all into your original equation to make sure that they're all valid.

**Zacken**)Yep. Just plug them all into your original equation to make sure that they're all valid.

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#16

(Original post by

Hi

The original question is: √2 × cos(x) = 1 / tan(x)

The book seems to have missed a possible answer.

**the81kid**)Hi

The original question is: √2 × cos(x) = 1 / tan(x)

The book seems to have missed a possible answer.

(Original post by

That's a good point. I should remember to do that.

**the81kid**)That's a good point. I should remember to do that.

**can't**have because you'd be dividing by zero in your original equation.

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(Original post by

In which case you'd find that you

**Zacken**)In which case you'd find that you

**can't**have because you'd be dividing by zero in your original equation.Am I understanding that correctly?

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#18

(Original post by

Aah. So the 1 / tan(x) part means that x can't / does not equal zero, BECAUSE they wouldn't / can't divide by 0 originally?

Am I understanding that correctly?

**the81kid**)Aah. So the 1 / tan(x) part means that x can't / does not equal zero, BECAUSE they wouldn't / can't divide by 0 originally?

Am I understanding that correctly?

So aren't valid as that would lead to you dividing by zero originally.

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(Original post by

It means that can't / does not equal zero, because you wouldn't / can't divide by 0 originally.

So aren't valid as that would lead to you dividing by zero originally.

**Zacken**)It means that can't / does not equal zero, because you wouldn't / can't divide by 0 originally.

So aren't valid as that would lead to you dividing by zero originally.

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#20

(Original post by

So the original problem/equation they give me, shows that x does/can not equal zero? Thanks, this helps me.

**the81kid**)So the original problem/equation they give me, shows that x does/can not equal zero? Thanks, this helps me.

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