giải PT a, sin(15độ -2x)=-1 b, sin2x+sin5x=0 c, cosx – 3 độ= $\frac{√2}{2}$ d, sin ²x +sin ²3x =1

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1. Đáp án:

   $\begin{array}{l}
   a)\sin \left( {{{15}^0} – 2x} \right) =  – 1\\
    \Rightarrow {15^0} – 2x =  – {90^0} + k{.360^0}\\
    \Rightarrow 2x = {105^0} – k{.360^0}\\
    \Rightarrow x = 52,{5^0} + k{.180^0}\\
   b)\sin 2x + \sin 5x = 0\\
    \Rightarrow sin2x =  – sin5x\\
    \Rightarrow sin2x = \sin \left( { – 5x} \right)\\
    \Rightarrow \left[ \begin{array}{l}
   2x =  – 5x + k2\pi \\
   2x = \pi  + 5x + k2\pi 
   \end{array} \right.\left( {k \in Z} \right)\\
    \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{{k2\pi }}{7}\\
   x = \dfrac{{ – \pi }}{3} – \dfrac{{k2\pi }}{3}
   \end{array} \right.\left( {k \in Z} \right)\\
   c)\cos \left( {x – {3^0}} \right) = \dfrac{{\sqrt 2 }}{2}\\
    \Rightarrow \cos \left( {x – {3^0}} \right) = \cos {45^0}\\
    \Rightarrow \left[ \begin{array}{l}
   x – {3^0} = {45^0} + k{.360^0}\\
   x – {3^0} =  – {45^0} + k{.360^0}
   \end{array} \right.\left( {k \in Z} \right)\\
    \Rightarrow \left[ \begin{array}{l}
   x = {48^0} + k{.360^0}\\
   x =  – {42^0} + k{.360^0}
   \end{array} \right.\left( {k \in Z} \right)\\
   d){\sin ^2}x + {\sin ^2}3x = 1\\
    \Rightarrow {\sin ^2}x + {\sin ^2}3x = {\sin ^2}x + {\cos ^2}x\\
    \Rightarrow {\sin ^2}3x = {\cos ^2}x\\
    \Rightarrow \left[ \begin{array}{l}
   \sin 3x = \cos x\\
   \sin 3x =  – \cos x
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   \sin 3x = \sin \left( {\dfrac{\pi }{2} – x} \right)\\
   \sin 3x = \sin \left( {x – \dfrac{\pi }{2}} \right)
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   3x = \dfrac{\pi }{2} – x + k2\pi \\
   3x = \pi  – \dfrac{\pi }{2} + x + k2\pi \\
   3x = x – \dfrac{\pi }{2} + k2\pi \\
   3x = \pi  – x + \dfrac{\pi }{2} + k2\pi 
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
   x = \dfrac{\pi }{4} + k\pi \\
   x =  – \dfrac{\pi }{4} + k\pi \\
   x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{{k\pi }}{4}\\
   x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
   x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}
   \end{array} \right.\left( {k \in Z} \right)
   \end{array}$

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