Toán cosx – √3 sinx = √2 3sin ²x + 5cos ²x – 2cos2x – 4sin2x = 0 05/08/2021 By Ariana cosx – √3 sinx = √2 3sin ²x + 5cos ²x – 2cos2x – 4sin2x = 0
Đáp án: Giải thích các bước giải: Phương trình tương đương :3sin2x+5cos2x−2(cos2x−sin2x)−8sinx.cosx=03sin2x+5cos2x−2(cos2x−sin2x)−8sinx.cosx=0 ⇔4sin2x+3cosx2−8sinx.cosx=0⇔4sin2x+3cosx2−8sinx.cosx=0 ⇔(2sinx−3cosx)(2sinx−cosx)=0 cosx + √3.sinx = √2<=>1/2cosx+√3/2 sinx=√2/2<=>cos π/3 cosx+sin π/3 sinx=√2/2<=>cos(x- π/3)=cos π/4<=>[x=7π/12+k2π [x=π/12+k2π Trả lời
Đáp án: $\begin{array}{l} + )\\\cos x – \sqrt 3 \sin x = \sqrt 2 \\ \Rightarrow \frac{1}{2}\cos x – \frac{{\sqrt 3 }}{2}\sin x = \frac{{\sqrt 2 }}{2}\\ \Rightarrow \cos \frac{\pi }{3}.\cos x – \sin \frac{\pi }{3}.\sin x = \frac{{\sqrt 2 }}{2}\\ \Rightarrow \cos \left( {x + \frac{\pi }{3}} \right) = \cos \frac{\pi }{4}\\ \Rightarrow \left[ \begin{array}{l}x + \frac{\pi }{3} = \frac{\pi }{4} + k2\pi \\x + \frac{\pi }{3} = – \frac{\pi }{4} + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = – \frac{\pi }{{12}} + k2\pi \\x = – \frac{{7\pi }}{{12}} + k2\pi \end{array} \right.\\ + )\\3{\sin ^2}x + 5{\cos ^2}x – 2\cos 2x – 4\sin 2x = 0\\ \Rightarrow 3{\sin ^2}x + 5{\cos ^2}x – 2\left( {{{\cos }^2}x – {{\sin }^2}x} \right) – 4\sin 2x = 0\\ \Rightarrow 5{\sin ^2}x + 3{\cos ^2}x – 4.2\sin x.\cos x = 0\\ \Rightarrow 5.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} – 8.\frac{{\sin x}}{{\cos x}} + 3 = 0\left( {do:{{\cos }^2}x \ne 0} \right)\\ \Rightarrow 5{\tan ^2}x – 8.\tan x + 3 = 0\\ \Rightarrow \left[ \begin{array}{l}\tan x = 1\\\tan x = \frac{3}{5}\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x = \arctan \frac{3}{5} + k\pi \end{array} \right.\end{array}$ Trả lời
Đáp án:
Giải thích các bước giải:
Phương trình tương đương :3sin2x+5cos2x−2(cos2x−sin2x)−8sinx.cosx=03sin2x+5cos2x−2(cos2x−sin2x)−8sinx.cosx=0
⇔4sin2x+3cosx2−8sinx.cosx=0⇔4sin2x+3cosx2−8sinx.cosx=0
⇔(2sinx−3cosx)(2sinx−cosx)=0
cosx + √3.sinx = √2
<=>1/2cosx+√3/2 sinx=√2/2
<=>cos π/3 cosx+sin π/3 sinx=√2/2
<=>cos(x- π/3)=cos π/4
<=>[x=7π/12+k2π
[x=π/12+k2π
Đáp án:
$\begin{array}{l}
+ )\\
\cos x – \sqrt 3 \sin x = \sqrt 2 \\
\Rightarrow \frac{1}{2}\cos x – \frac{{\sqrt 3 }}{2}\sin x = \frac{{\sqrt 2 }}{2}\\
\Rightarrow \cos \frac{\pi }{3}.\cos x – \sin \frac{\pi }{3}.\sin x = \frac{{\sqrt 2 }}{2}\\
\Rightarrow \cos \left( {x + \frac{\pi }{3}} \right) = \cos \frac{\pi }{4}\\
\Rightarrow \left[ \begin{array}{l}
x + \frac{\pi }{3} = \frac{\pi }{4} + k2\pi \\
x + \frac{\pi }{3} = – \frac{\pi }{4} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – \frac{\pi }{{12}} + k2\pi \\
x = – \frac{{7\pi }}{{12}} + k2\pi
\end{array} \right.\\
+ )\\
3{\sin ^2}x + 5{\cos ^2}x – 2\cos 2x – 4\sin 2x = 0\\
\Rightarrow 3{\sin ^2}x + 5{\cos ^2}x – 2\left( {{{\cos }^2}x – {{\sin }^2}x} \right) – 4\sin 2x = 0\\
\Rightarrow 5{\sin ^2}x + 3{\cos ^2}x – 4.2\sin x.\cos x = 0\\
\Rightarrow 5.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} – 8.\frac{{\sin x}}{{\cos x}} + 3 = 0\left( {do:{{\cos }^2}x \ne 0} \right)\\
\Rightarrow 5{\tan ^2}x – 8.\tan x + 3 = 0\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = \frac{3}{5}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + k\pi \\
x = \arctan \frac{3}{5} + k\pi
\end{array} \right.
\end{array}$