Toán 3sin^2x +sin2x+cos^2x = 3 ( kí hiệu: ^ là mũ nha) 16/09/2021 By Faith 3sin^2x +sin2x+cos^2x = 3 ( kí hiệu: ^ là mũ nha)
$3{\sin}^2x +\sin2x+{\cos}^2x = 3 $ $\Rightarrow 3\dfrac{1-\cos 2x}{2}+\sin 2x+\dfrac{\cos 2x+1}{2}=3$ $\Rightarrow -\cos 2x+\sin 2x=1$ $\Rightarrow \sqrt2\sin \left({2x-\dfrac{\pi}{4}}\right)=1$ $\Rightarrow \sin \left({2x-\dfrac{\pi}{4}}\right)=\dfrac{1}{\sqrt2} $ $\Rightarrow \left[ \begin{array}{l} 2x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi \\ 2x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi \end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} x=\dfrac{\pi}{2}+k\pi \\ x=\dfrac{\pi}{2}+k\pi \end{array} \right .(k\in\mathbb Z)$ Trả lời
Đáp án: Giải thích các bước giải: 3.(1-cos2x)/2+sin2x+(1+cos2x)/2=3 <=>3-3cos2x+2sin2x+1+cos2x=6 <=>-2cos2x+2sin2x+4=6 <=>cos2x-sin2x=-1 <=>(1/căn2).cos2x-(1/căn2).sin2x=-1/căn2 <=>sin(pi/4-2x)=sin(-pi/4) bạn giải tiếp nhé Trả lời
$3{\sin}^2x +\sin2x+{\cos}^2x = 3 $
$\Rightarrow 3\dfrac{1-\cos 2x}{2}+\sin 2x+\dfrac{\cos 2x+1}{2}=3$
$\Rightarrow -\cos 2x+\sin 2x=1$
$\Rightarrow \sqrt2\sin \left({2x-\dfrac{\pi}{4}}\right)=1$
$\Rightarrow \sin \left({2x-\dfrac{\pi}{4}}\right)=\dfrac{1}{\sqrt2} $
$\Rightarrow \left[ \begin{array}{l} 2x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi \\ 2x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi \end{array} \right .$
$\Rightarrow \left[ \begin{array}{l} x=\dfrac{\pi}{2}+k\pi \\ x=\dfrac{\pi}{2}+k\pi \end{array} \right .(k\in\mathbb Z)$
Đáp án:
Giải thích các bước giải: 3.(1-cos2x)/2+sin2x+(1+cos2x)/2=3
<=>3-3cos2x+2sin2x+1+cos2x=6
<=>-2cos2x+2sin2x+4=6
<=>cos2x-sin2x=-1
<=>(1/căn2).cos2x-(1/căn2).sin2x=-1/căn2
<=>sin(pi/4-2x)=sin(-pi/4)
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