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## #1 2013-05-26 20:53:09

atuturay
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### Roots of an Equation

Hiii,

I don't know how to work out the root of the equation sinx= 1+cosx for 0°<x<180°

## #2 2013-05-26 21:16:20

zetafunc.
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### Re: Roots of an Equation

Squaring both sides might help.

## #3 2013-05-26 21:40:11

{7/3}
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### Re: Roots of an Equation

Bring cos x to left side and squre.

There are 10 kinds of people in the world,people who understand binary and people who don't.

## #4 2013-05-26 22:18:10

bob bundy
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### Re: Roots of an Equation

hi atuturay

Welcome to the forum.

Squaring will give the solutions.  It may also generate values that don't fit the original because (-z)^2 = z^2

An alternative starts like this:

Solve this firstly for x plus alpha and hence get x.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #5 2013-05-26 22:23:27

zetafunc.
Guest

### Re: Roots of an Equation

Letting x = arccos(u) or arcsin(u) is another way, although the method is mostly the same as in posts #2 and #3.

## #6 2013-05-26 23:31:22

anonimnystefy
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### Re: Roots of an Equation

Won't squaring introduce extraneous solutions?

The limit operator is just an excuse for doing something you know you can't.
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“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #7 2013-05-26 23:54:59

bob bundy
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### Re: Roots of an Equation

Won't squaring introduce extraneous solutions?

#### In post 4 I wrote:

Squaring will give the solutions.  It may also generate values that don't fit the original because (-z)^2 = z^2

But, in this example, as we only want 0 < x < 180, the single, correct solution may be found by squaring.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei