Đáp án:
Giải thích các bước giải:
a) $sin2x + cos2x + tanx = 2 $
$<=> 2sinx .cosx + 1- 2sin^2x + tanx = 2 $
$<=> 2 sinx.cosx - 1-2sin^2x + tanx = 0$
$<=> sin2x . cosx - cosx - 2sin^2x . cosx + sinx =0$
$<=> sin2x . cosx - sin2x . sinx -cosx + sinx = 0$
$<=> sin2x (cosx - sinx) - (cosx- sinx) = 0$
$<=> (sin2x - 1 )( cosx - sinx) = 0$
<=> \(\left[ \begin{array}{l}sin2x = 1 \\cosx = sin x\end{array} \right.\) <=> $x = \frac{pi}{4} + kpi$ $(k∈Z)$
b)$\frac{sinx + sin2x + sin3x}{cosx + cos2x + cos3x} = \sqrt{3}$
$<=> \frac{2sin2x . cosx + sin2x}{2cos2x.cosx + cos2x} = \sqrt{3}$
$<=> \frac{sin2x (2cosx + 1)}{cos2x(2cosx + 1)}$ = \sqrt{3}$
$<=> tan2x = \sqrt{3}$
$<=> 2x = \frac{pi}{3} + kpi$
$<=> x= \frac{pi}{6} + \frac{kpi}{2}$ $(k ∈ Z)$
