Câu hỏi:
Giúp em bài tập về nhà Toán lớp 11 câu hỏi như sau: a. Sin3x + cos2x – sinx = 0
b. Sin5x + 2cos^2x = 1
c. Sinx + 4cosx = 2 + sin2x
d. Cos3x + cos2x – cosx – 1 = 0
e. Sin^2 (x – π/4) = cos^2 x
f. 5cosx – sin2x = 0
Giúp em với ạ ???? em cảm ơn nhiều ạ
Trả lời 2:
Gia Sư Hoàng Khang gữi câu trả lời dành cho bạn:
a)
sin3x +cos2x -sinx=0
<=>(sin3x-sinx)+cos2x=0
<=>2cos2x.sin x+cos2x=0
cos2x(2sinx+1)=0
cos2x=0; 2x=π/2+kπ
x=π/4+kπ!2
2sinx+1=0;
x=-π/6+k2π
x=5π/6+k2π
b)
sin5x + 2cos^2x = 1
<=>sin5x =1-2cos^2x=-cos2x=cos(-2x)=sin[(π/2-(-2x)]
=sin(2x+π/2)
5x=2x,+π/2+k2π; x=π/6+2kπ/3
5x=-2x+π/2+k2π; x=π/14+k2π/7
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c)
sinx + 4cosx = 2 + sin2x
<=> sinx + 4cosx = 2 + 2sin xcosx
<=> [sinx – 2sinx cosx] + 4cosx -2 =0
<=> sinx(1 – 2 cosx] -2(1-2cosx) =0
<=> (1 – 2 cosx)(sinx -2) =0
<=> (1 – 2 cosx) =0
cosx =1/2
x =+-pi /3 + k2pi
d)
cos3x+cos2x-cosx-1=0
>>4cos^3(x)-3cosx+2cos^2(x)-1-cosx-1=0
>>4cos^3(x)+2cos^2(x)-4cosx-2=0
>>2cos^3(x)+cos^2(x)-2cosx-1=0
>>cos^2(x)(2cosx+1)-(2cosx+1)=0
>>(2cosx+1)(cos^2(x)-1)=0
>>(2cosx+1)(cosx-1)(cosx+1)=0
>>2cosx+1=0; cosx-1=0 hoặc cosx+1=0
Trả lời 1:
Gia Sư Hoàng Khang gữi câu trả lời dành cho bạn:
Giải đáp:
a)$\left[ \begin{array}{l}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7}{6}\pi+k2\pi\end{array} \right.$
b)$ \left[ \begin{array}{l}x=-\dfrac{\pi}{6}-\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{14}-\dfrac{k2\pi}{7}\end{array} \right.$
c)$\left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{array} \right.$
d)$\left[ \begin{array}{l}x=\dfrac{2\pi}{3}+k2\pi\\x=k2\pi\\x=\pi+k2\pi\end{array} \right. $
e)$ x=\dfrac{3\pi}{8}+k\pi$
f)$ x=\dfrac{\pi}{2}+k\pi$
Lời giải và giải thích chi tiết:
a) $sin 3x+cos 2x-sin x=0$
$\Rightarrow (sin 3x-sin x)+cos 2x=0$
$\Rightarrow 2cos2x.sinx +cos2x=0$
$\Rightarrow cos 2x.(2sin x+1)=0$
$\Rightarrow \left[ \begin{array}{l}cos 2x=0\\2sin x+1=0\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}cos 2x=0\\sin x=-\dfrac{1}{2}\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}2x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7}{6}\pi+k2\pi\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7}{6}\pi+k2\pi\end{array} \right.$
b) $sin 5x+2cos^2x=1$
$\Rightarrow sin5x+2cos^2x-1=0$
$\Rightarrow sin5x+cos2x=0$
$\Rightarrow -sin5x=cos2x$
$\Rightarrow sin(-5x)=sin(\dfrac{\pi}{2}-2x)$
$\Rightarrow \left[ \begin{array}{l}-5x=\dfrac{\pi}{2}-2x+k2\pi\\-5x=\pi-\dfrac{\pi}{2}+2x+k2\pi\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}x=-\dfrac{\pi}{6}-\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{14}-\dfrac{k2\pi}{7}\end{array} \right.$
c) $sinx+4cosx=2+sin2x$
$\Rightarrow sinx+4cosx-2-2sinx.cosx=0$
$\Rightarrow sinx(1-2cosx)+2(2cosx-1)=0$
$\Rightarrow (2cosx-1)(2-sinx)=0$
$\Rightarrow \left[ \begin{array}{l}2cosx-1=0\\2-sinx=0\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}cosx=\dfrac{1}{2}\\sinx=2 (L)\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{array} \right.$
d) $cos3x +cos2x-cosx-1=0$
$4cos^3x-3cosx+2cos^2x-1-cosx-1=0$
$\Rightarrow 4cos^3x+2cos^2x-4cosx-2=0$
$\Rightarrow 2cos^3x+cos^2x-2cosx-1=0$
$\Rightarrow cos^2x(2cosx+1)-(2cosx+1)=0$
$\Rightarrow (2cosx+1)(cos^2x-1)=0$
$\Rightarrow (2cosx+1)(cosx-1)(cosx+1)=0$
$\Rightarrow \left[ \begin{array}{l}2cosx+1=0\\cosx-1=0\\cosx+1=0\end{array} \right. $
$\Rightarrow \left[ \begin{array}{l}x=\dfrac{2\pi}{3}+k2\pi\\x=k2\pi\\x=\pi+k2\pi\end{array} \right. $
e) $sin^2(x-\dfrac{\pi}{4})=cos^2x$
$\Rightarrow \left[ \begin{array}{l}sin(x-\dfrac{\pi}{4})=cosx\\sin(x-\dfrac{\pi}{4})=-cosx\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{\pi}{2}-x+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{\pi}{2}+x+k2\pi\end{array} \right.$
$\Rightarrow x=\dfrac{3\pi}{8}+k\pi$
f) $5cosx-sin2x=0$
$\Rightarrow 5cosx-2sinx.cosx=0$
$\Rightarrow cosx(5-2sinx)=0$
$\Rightarrow \left[ \begin{array}{l}cosx=0\\sinx=\dfrac{5}{2}(L)\end{array} \right.$
$\Rightarrow x=\dfrac{\pi}{2}+k\pi$
