Đáp án:
a) 2sinx - m = 0
⇔ sinx = $\frac{m}{2}$
Pt có nghiệm khi:
-1≤$\frac{m}{2}$ ≤1
⇔ -2≤m≤2
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2) giải pt :
√3 sinx - cosx =2
⇔ $\frac{√3}{2}$ sinx - $\frac{1}{2}$cosx =1
⇔ cos$\frac{π}{6}$.sinx - sin$\frac{π}{6}$.cosx = 1
⇔ sin ($\frac{π}{6}$ -x) = 1
⇔ sin ($\frac{π}{6}$-x) = sin$\frac{π}{2}$
⇔\(\left[ \begin{array}{l}π/6 - x=π/2 + k2 π \\π/6 - x= π- π/2 +k2 π\end{array} \right.\) ⇔ \(\left[ \begin{array}{l} x=2π/3 + k2 π \\ x= 2π/3 + k2 π\end{array} \right.\) (k∈Z)
Vậy x = 2π /3 +k2π
