sin(inverse)x+sin(inverse)2x=pi/3

Question

solve for x

Answer 1

This is a good one Pavan.  It had me going down a few rabbit trails before I found the correct path (as trig functions sometimes do.)

 

sin^-1(x)+sin^-1(2x)=π/3

 

First, it is helpful if we rewrite this expression parametrically:

a+b=π/3

a=sin^-1(x)

b=sin^-1(2x)

 

In turn we can solve the second and third equations for x:

x=sin(a)

2x=sin(b)

 

And then substitute sin(a) into 2x=sin(b) for:

2sin(a)=sin(b)

 

Now we have a system of two equations with two variable we can solve:

a+b=π/3

2sin(a)=sin(b)

 

Isolate a in the first equation:

a=π/3-b

 

And substitute:

2sin(π/3-b)=sin(b)

 

Here is the tricky part.  We have to utilize the trig identity that says:

sin(α-β)=sin(α)cos(β)-cos(α)sin(β) where we will use α=π/3 and β=b.

 

This will change our expression to:

2(sin(π/3)cos(b)-cos(π/3)sin(b)=sin(b)

2((√3/2)cos(b)-(1/2)sin(b)=sin(b)               Simplify

√3cos(b)-sin(b)=sin(b)                              Simplify

√3cos(b)=2sin(b)                                     add sin(b) to both sides

√3=2sin(b)/cos(b)                                    divide cos(b)

√3/2=sin(b)/cos(b)                                   divide 2

tan(b)=√3/2                                            tan=sin/cos

b=tan^-1(√3/2)                                          Convert to inverse

b≈.713724379

 

Since we already know that 2x=sin(b), we plug it in and get x≈.327326835.