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

Question

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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3 tháng 2021-09-20T01:00:04+00:00 1 Answers 4 views 0

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