Question 881000
<pre>
sin(4x)cos(3x)+cos(4x)sin(3x)=0.51

Use the identity

{{{sin(alpha+beta)}}} {{{""=""}}} {{{sin(alpha)*cos(beta)}}}{{{""+""}}}{{{cos(alpha)*sin(beta)}}}

Substitute {{{alpha=4x}}}, {{{beta=3x}}}:

{{{sin(4x+3x)}}} {{{""=""}}} {{{sin(4x)*cos(3x)}}}{{{""+""}}}{{{cos(4x)*sin(3x)}}}

or

{{{sin(7x)}}} {{{""=""}}} {{{sin(4x)*cos(3x)}}}{{{""+""}}}{{{cos(4x)*sin(3x)}}}

{{{sin(7x)}}} {{{""=""}}} {{{0.51}}}

The right side equals to 0.51.  

Take inverse sines of both sides:

{{{7x}}} {{{""=""}}} sin<sup>-1</sup>(0.51) + {{{2pi*n}}}

also

{{{7x}}} {{{""=""}}} ({{{pi}}} - sin<sup>-1</sup>(0.51) + {{{2n*pi}}})

{{{7x}}} {{{""=""}}} -sin<sup>-1</sup>(0.51) + {{{(2n+1)pi)}}})
Therefore, solving for x in each

{{{x}}} {{{""=""}}} {{{1/7}}}sin<sup>-1</sup>(0.51) + {{{2pi*n/7}}}

also

{{{x}}} {{{""=""}}} {{{-1/7}}}sin<sup>-1</sup>(0.51) + {{{(2n+1)pi/7)}}})

To get the two fundamental solutions let n=0 in each:


{{{x}}} {{{""=""}}} {{{1/7}}}sin<sup>-1</sup>(0.51)

and


{{{x}}} {{{""=""}}} {{{-1/7}}}sin<sup>-1</sup>(0.51) + {{{pi/7)}}}

Edwin</pre>