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**atuturay****Member**- Registered: 2013-05-25
- Posts: 1

Hiii,

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

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**zetafunc.****Guest**

Squaring both sides might help.

**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

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.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,020

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

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**zetafunc.****Guest**

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

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

Won't squaring introduce extraneous solutions?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,020

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

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