1. sin3x = sinx
⇔ \(\left[ \begin{array}{l}3x=x+k2π\\3x=π-x+k2π\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x=k2π\\4x=π+k2π\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=kπ\\x=π/4+kπ/2\end{array} \right.\)
⇔ x=π/4 +kπ/2 (k∈Z)
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2. sinx – cos2x =0
⇔ sinx = cos2x
⇔ sinx= sin(π/2 - x)
⇔ \(\left[ \begin{array}{l}x=π/2-x+k2π\\x=π-π/2+x+k2π\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=π/4+kπ(k∈Z)\\x=π-π/2+x+k2π(loại)\end{array} \right.\)
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3. sin2x = cos3x
⇔ sin2x = sin(π/2 - 3x)
⇔ \(\left[ \begin{array}{l}2x=π/3-3x+k2π\\2x=π-π/2+3x+k2π\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=π/10+k2π/5\\x=-π/2-k2π\end{array} \right.\) (k∈Z)
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4. sin4x + cos5x=0
⇔ cos5x = sin(-4x)
⇔ cos5x = sin(π/2 +4x)
⇔ \(\left[ \begin{array}{l}5x=π/2+4x+k2π\\5x=-π/2-4x+k2π\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=π/2+k2π\\x=-π/18+k2π/9\end{array} \right.\) (k∈Z)
