Question

sinx+3sin2x+sin3x=cosx+3cos2x+cos3x , where x=(0,2π)
The difference between the greatest and the least is​

Answer 1

Answer:

$\textbf{using C-D formulae :}$

we get

$\text{2sin2xcosx−3sin2x=2cos2xcosx−3cos2x}$

$\text{(2cosx−3)sin2x=(2cosx−3)cos2x}$

$\text{cos2x=sin2x}$

$\text{tan2x=1}$

$x = \frac{nΠ}{2} + \frac{nΠ}{8}$

$\\ \\ \\ \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙ}$