sinx+ sin2x+ sin3x= 1
thanks for all
Post #2 puts you on the right track. I'll add that line 2 of that reply follows from using the compound angle formula on sin(x + 2x).
You need to show some effort if more help is needed. What have you tried and where are you still stuck? Are you expected to solve it by hand or using CAS technology?
Not the whole solution...
With help of my formula we get:
$\displaystyle \sin2x\sin{\frac{3x}{2}}=\sin{\frac{x}{2}}$
$\displaystyle 2\sin{x}\cos{x}\sin{\frac{3x}{2}}=\sin{\frac{x}{2} }$
$\displaystyle 4\sin {\frac{x}{2}} \cos{\frac{x}{2}} \cos{x} \sin {\frac{3x}{2}}=\sin{\frac{x}{2}}$
$\displaystyle \sin {\frac{x}{2}} (4\cos {\frac{x}{2}} \cos {x} \sin {\frac{3x}{2}}-1)=0$
$\displaystyle \sin {\frac{x}{2}} (4\cos {\frac{x}{2}}\cos {x} \sin {\frac{3x}{2}}-1)=0$
$\displaystyle \sin {\frac{x}{2}} (2\sin {2x}+\sin {2x}-1)=0$
$\displaystyle \sin {\frac{x}{2}} (2 \sin {2x} \cos {x}+\sin {2x}-1)=0$
This equation does not seem to have a clean solution. If you graph the function, you can see that there are two solutions between 0º and 90º (and no others in the range 0º to 360º). The smaller solution is approximately 9.836º, but that does not look like a recognisable quantity in either degrees or radians. I haven't tried to locate the other solution accurately, but it is somewhere near 71.3º. Again, this is not a nice-looking angle. The solutions for $\displaystyle \sin x$ are not rational. I doubt whether the equation can be solved exactly.