Toán Giải pt lượng giác sau √3 cosx – sinx= √2 18/07/2021 By Caroline Giải pt lượng giác sau √3 cosx – sinx= √2
`\sqrt{3} cos(x) – sin(x) = \sqrt{2}` `⇔ \sqrt{3} cos(x) = \sqrt{2} + sin(x)` `⇔ 3cos^2(x) – 2 – 2\sqrt{2} sin(x) – sin^2(x) = 0` `⇔ 1 – 4sin^2(x) – 2sin(x)\sqrt{2} = 0` `⇔ sin(x) = -(\sqrt{2}+\sqrt{6})/4 , sin(x) = (\sqrt{6}-\sqrt{2})/4` `⇔ x = arcsin(-(\sqrt{2}+\sqrt{6})/4) + 2πn , x=arcsin((\sqrt{6}-\sqrt{2})/4) + 2πn` Trả lời
Giải thích các bước giải: `sqrt3 cosx – sinx= sqrt2` `=>sqrt3/2cosx+1/2sinx=sqrt2/2` `=>sin(pi/3-2x)=sin“pi/3` `=>`\(\left[ \begin{array}{l}\dfrac{\pi}3-2x=\dfrac{\pi}3+k2\pi\\\dfrac{\pi}3-2x=\pi-\dfrac{\pi}3+k2\pi\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=k\pi\\x=-\dfrac{\pi}6+k\pi\end{array} \right.\text{ }(k\in \Bbb Z)\) Trả lời
`\sqrt{3} cos(x) – sin(x) = \sqrt{2}`
`⇔ \sqrt{3} cos(x) = \sqrt{2} + sin(x)`
`⇔ 3cos^2(x) – 2 – 2\sqrt{2} sin(x) – sin^2(x) = 0`
`⇔ 1 – 4sin^2(x) – 2sin(x)\sqrt{2} = 0`
`⇔ sin(x) = -(\sqrt{2}+\sqrt{6})/4 , sin(x) = (\sqrt{6}-\sqrt{2})/4`
`⇔ x = arcsin(-(\sqrt{2}+\sqrt{6})/4) + 2πn , x=arcsin((\sqrt{6}-\sqrt{2})/4) + 2πn`
Giải thích các bước giải:
`sqrt3 cosx – sinx= sqrt2`
`=>sqrt3/2cosx+1/2sinx=sqrt2/2`
`=>sin(pi/3-2x)=sin“pi/3`
`=>`\(\left[ \begin{array}{l}\dfrac{\pi}3-2x=\dfrac{\pi}3+k2\pi\\\dfrac{\pi}3-2x=\pi-\dfrac{\pi}3+k2\pi\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=k\pi\\x=-\dfrac{\pi}6+k\pi\end{array} \right.\text{ }(k\in \Bbb Z)\)