Math Is Fun Forum

atuturay

Member

Registered: 2013-05-25

Posts: 1

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°

zetafunc.

Guest

Re: Roots of an Equation

Squaring both sides might help.

{7/3}

Member

Registered: 2013-02-11

Posts: 210

Re: Roots of an Equation

Bring cos x to left side and squre.

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,644

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

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

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.

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Roots of an Equation

Won't squaring introduce extraneous solutions?

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,644

Re: Roots of an Equation

Won't squaring introduce extraneous solutions?

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

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob