1)
a) √3 sin2x+sin(π/2 +2x) =1
b.cos2x+3cosx+2=0
c) sin3x+cos3x-sinx+cosx=√2 cos2x
d)9-13cosx+4/1+tan^2x=0
1) a) √3 sin2x+sin(π/2 +2x) =1 b.cos2x+3cosx+2=0 c) sin3x+cos3x-sinx+cosx=√2 cos2x d)9-13cosx+4/1+tan^2x=0
By Liliana
By Liliana
1)
a) √3 sin2x+sin(π/2 +2x) =1
b.cos2x+3cosx+2=0
c) sin3x+cos3x-sinx+cosx=√2 cos2x
d)9-13cosx+4/1+tan^2x=0
a, Ta có ` sqrt3.sin2x+sin(π/2 +2x) =1`
`<=> cos(2 x) + sqrt(3) sin(2 x) = 2 [1/2 cos(2 x) + 1/2 sqrt(3) sin(2 x)]`
`= 2 [sin(π/6) cos(2 x) + cos(π/6) sin(2 x)]`
`= 2 sin(2 x + π/6): 2 sin(2 x + π/6) = 1`
`<=> sin(2 x + π/6) = 1/2`
`<=>` \(\left[ \begin{array}{l}2 x + \frac{\pi}{6} = 2 π n + \frac{5\pi}{6} (n \in \mathbb{Z})\\2 x + \frac{\pi}{6} = 2 π n + \frac{\pi}{6}(n \in \mathbb{Z})\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2 x = 2 π n + \frac{2\pi}{3}(n \in \mathbb{Z})\\2 x = 2 π n (n \in \mathbb{Z})\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = π n + \frac{\pi}{3}(n \in \mathbb{Z})\\ x = π n (n \in \mathbb{Z}) \end{array} \right.\)
Vậy phương trình có tập nghiệm
`S={π n +\frac{\pi}{3} ; π n | n \in \mathbb{Z}}`
b) Ta có: `cos2x+3cosx+2=0`
`<=> 1 + 3 cos(x) + 2 cos^2(x) = 0`
`<=> [cos(x) + 1] [2 cos(x) + 1] = 0`
`<=>` \(\left[ \begin{array}{l}cos(x) + 1 = 0\\ 2 cos(x) + 1 = 0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}cos(x) = -1\\ 2 cos(x) = -1 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = 2 π n + π (n \in \mathbb{Z})\\ cos(x) = \frac{-1}{2} \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = 2 π n + π (n \in \mathbb{Z})\\ x = 2 π n + \frac{2\pi}{3}(n \in \mathbb{Z})\\x = 2 π n + \frac{4\pi}{3}(n \in \mathbb{Z})\end{array} \right.\)
Vậy phương trình có tập nghiệm:
`S={2 π n + π ; 2 π n + \frac{2\pi}{3} ; x = 2 π n + \frac{4\pi}{3} | n \in \mathbb{Z}}`
c, Ta có: `sin3x+cos3x-sinx+cosx=\sqrt2 cos2x`
`<=> cos(x) – sqrt(2) cos(2 x) + cos(3 x) – sin(x) + sin(3 x) = 0`
`<=> 2 sin(-x + π/4) sin(x + π/4) [2 sqrt(2) sin(x + π/4) – sqrt(2)] = 0`
`<=> sin(-x + π/4) sin(x + π/4) [2 sqrt(2) sin(x + π/4) – sqrt(2)] = 0`
`<=>` \(\left[ \begin{array}{l}sin(-x + \frac{\pi}{4}) = 0\\ sin(x + \frac{\pi}{4}) = 0\\2 \sqrt2 sin(x + \frac{\pi}{4}) -\sqrt2 = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}-x + \frac{\pi}{4} = π n(n\in \mathbb{Z})\\ x + \frac{\pi}{4} = π n (n\in \mathbb{Z})\\\sqrt2 [2 sin(x + \frac{\pi}{4}) – 1] = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} – π n(n\in \mathbb{Z})\\ x = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\2 sin(x + \frac{\pi}{4}) – 1 = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} – π n(n\in \mathbb{Z})\\ x = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\ sin(x + \frac{\pi}{4}) = \frac{1}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} – π n(n\in \mathbb{Z})\\ x = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\ x + \frac{\pi}{4}= 2 π n + \frac{5\pi}{6}(n\in \mathbb{Z})\\x + \frac{\pi}{4}= 2 π n + \frac{\pi}{6}(n\in \mathbb{Z})\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} – π n(n\in \mathbb{Z})\\ x = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\ x = 2 π n + \frac{7\pi}{12}(n\in \mathbb{Z})\\x = 2 π n – \frac{\pi}{12}(n\in \mathbb{Z})\end{array} \right.\)
Vậy phương trình có tập nghiệm:
`S={\frac{\pi}{4} – π n ; π n-\frac{\pi}{4} ; x = 2 π n + \frac{7\pi}{12} ; 2 π n – \frac{\pi}{12} | n \in \mathbb{Z}}`
d) Đề bài là `(9-13cosx+4)/(1+tan^2x)=0` hay là như thế nào hả bạn? Sau mình bổ sung nhé.
Đáp án:
a)$x = \dfrac{\pi }{3} + k\pi ;x = k\pi \left( {k \in Z} \right)$
b)$x = \pi + k2\pi ;x = \dfrac{{2\pi }}{3} + k2\pi ;x = \dfrac{{ – 2\pi }}{3} + k2\pi \left( {k \in Z} \right)$
c)$x = \pm \dfrac{\pi }{4} + k\pi ;x = \dfrac{{7\pi }}{{12}} + k2\pi ;x = \dfrac{{ – \pi }}{{12}} + k2\pi \left( {k \in Z} \right)$
d)$x = k2\pi \left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
a)\sqrt 3 \sin 2x + \sin \left( {\dfrac{\pi }{2} + 2x} \right) = 1\\
\Leftrightarrow \sqrt 3 \sin 2x + \cos 2x = 1\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 2x + \dfrac{1}{2}\cos 2x = \dfrac{1}{2}\\
\Leftrightarrow \cos \left( {2x – \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
2x – \dfrac{\pi }{3} = \dfrac{{ – \pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
Vậy các họ nghiệm của phương trình là: $x = \dfrac{\pi }{3} + k\pi ;x = k\pi \left( {k \in Z} \right)$
$\begin{array}{l}
b)\cos 2x + 3\cos x + 2 = 0\\
\Leftrightarrow 2{\cos ^2}x + 3\cos x + 1 = 0\\
\Leftrightarrow \left( {\cos x + 1} \right)\left( {2\cos x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = – 1\\
\cos x = \dfrac{{ – 1}}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pi + k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = \dfrac{{ – 2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
Vậy các họ nghiệm của phương trình là:$x = \pi + k2\pi ;x = \dfrac{{2\pi }}{3} + k2\pi ;x = \dfrac{{ – 2\pi }}{3} + k2\pi \left( {k \in Z} \right)$
$\begin{array}{l}
c)\sin 3x + \cos 3x – \sin x + \cos x = \sqrt 2 \cos 2x\\
\Leftrightarrow 3\sin x – 4{\sin ^3}x + 4{\cos ^3}x – 3\cos x – \sin x + \cos x = \sqrt 2 \left( {{{\cos }^2}x – {{\sin }^2}x} \right)\\
\Leftrightarrow 2\left( {\sin x – \cos x} \right) – 4\left( {{{\sin }^3}x – {{\cos }^3}x} \right) + \sqrt 2 \left( {{{\sin }^2}x – {{\cos }^2}x} \right) = 0\\
\Leftrightarrow \left( {\sin x – \cos x} \right)\left( {2 – 4\left( {{{\sin }^2}x + \sin x\cos x + {{\cos }^2}x} \right) + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
\Leftrightarrow \left( {\sin x – \cos x} \right)\left( {2 – 4\left( {1 + \sin x\cos x} \right) + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
\Leftrightarrow \left( {\sin x – \cos x} \right)\left( { – 2 – 4\sin x\cos x + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
\Leftrightarrow \left( {\sin x – \cos x} \right)\left( { – 2{{\left( {\sin x + \cos x} \right)}^2} + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
\Leftrightarrow \left( {\sin x – \cos x} \right)\left( {\sin x + \cos x} \right)\left( { – 2\left( {\sin x + \cos x} \right) + \sqrt 2 } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x – \cos x = 0\\
\sin x + \cos x = 0\\
\sin x + \cos x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = – 1\\
\cos \left( {x – \dfrac{\pi }{4}} \right) = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{4} + k\pi \\
x – \dfrac{\pi }{4} = \dfrac{\pi }{3} + k2\pi \\
x – \dfrac{\pi }{4} = – \dfrac{\pi }{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k2\pi \\
x = \dfrac{{ – \pi }}{{12}} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
Vậy các họ nghiệm của phương trình là:$x = \pm \dfrac{\pi }{4} + k\pi ;x = \dfrac{{7\pi }}{{12}} + k2\pi ;x = \dfrac{{ – \pi }}{{12}} + k2\pi \left( {k \in Z} \right)$
$\begin{array}{l}
d)9 – 13\cos x + \dfrac{4}{{1 + {{\tan }^2}x}} = 0\left( {DK:x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)} \right)\\
\Leftrightarrow 9 – 13\cos x + 4{\cos ^2}x = 0\\
\Leftrightarrow \left( {\cos x – 1} \right)\left( {4\cos x – 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{9}{4}\left( l \right)
\end{array} \right.\\
\Leftrightarrow x = k2\pi \left( {k \in Z} \right)\left( {tm} \right)
\end{array}$
Vậy các họ nghiệm của phương trình là:$x = k2\pi \left( {k \in Z} \right)$