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Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra f'(x)=0
a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0
d,f(x)=\(\frac{3}{2}\)=>f'(x)=0
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)
⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)
⇔ 2cos2x - 5cosx + 2 = 0
⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên
2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)
⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0
⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)
⇒ sin4x + cos4x = 48.sin4x . cos4x
⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x
⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)
⇔ 1 - 2sin22x = 0
⇔ cos4x = 0
⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)
⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)
⇔ sin2x - sin22x - (1 + cos4x) = 0
⇔ sin2x - sin22x - 2cos22x = 0
⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0
⇔ sin22x + sin2x - 2 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)
⇔ sin2x = 1
⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
4, cos5x + cos2x + 2sin3x . sin2x = 0
⇔ cos5x + cos2x + cosx - cos5x = 0
⇔ cos2x + cosx = 0
⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
⇔ \(cos\dfrac{3x}{2}=0\)
⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)
⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)
Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)
⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}
Vậy các nghiệm thỏa mãn là các phần tử của tập hợp
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}cos4x+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}\left(1-2sin^22x\right)+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow...\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x+cos\left(x-\frac{\pi}{4}\right)sin\left(3x-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
Đặt \(x-\frac{\pi}{4}=a\Rightarrow x=a+\frac{\pi}{4}\)
\(\Rightarrow1-\frac{1}{2}sin^2\left(2a+\frac{\pi}{2}\right)+cosa.sin\left(3a+\frac{3\pi}{4}-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
\(\Leftrightarrow1-\frac{1}{2}cos^22a+cosa.cos3a-\frac{3}{2}=0\)
\(\Leftrightarrow2-cos^22a+cos4a+cos2a-3=0\)
\(\Leftrightarrow-cos^22a+2cos^22a-1+cos2a-1=0\)
\(\Leftrightarrow cos^22a+cos2a-2=0\)
\(\Leftrightarrow cos2a=1\Leftrightarrow cos\left(2x-\frac{\pi}{2}\right)=1\)
\(\Leftrightarrow sin2x=1\Rightarrow x=\frac{\pi}{4}+k\pi\)
a.
Tổng là cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-sin^2x\end{matrix}\right.\)
Do đó: \(S=\dfrac{u_1}{1-q}=\dfrac{1}{1+sin^2x}\)
b. Tương tự, tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=cos^2x\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{1}{1-cos^2x}=\dfrac{1}{sin^2x}\)
c. Do \(0< x< \dfrac{\pi}{4}\Rightarrow0< tanx< 1\)
Tổng trên vẫn là tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-tanx\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{1}{1+tanx}\)
4.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=cos2x\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=cos2x\)
\(\Leftrightarrow1+1-sin^22x=2cos2x\)
\(\Leftrightarrow1+cos^22x=2cos2x\)
\(\Leftrightarrow\left(cos2x-1\right)^2=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow2x=k2\pi\)
\(\Rightarrow x=k\pi\)
3.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{1}{2}\)
\(\Leftrightarrow1-sin^22x=0\)
\(\Leftrightarrow cos^22x=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
Lần sau bạn lưu ý viết đầy đủ đề bài.
Lời giải:
Đặt $|\sin x|=a; |\cos x|=b$ thì bài toán trở thành:
\(\left\{\begin{matrix} a^4-b^4=a+b\\ a^2+b^2=1\\ a,b\in [0;1]\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} a^2-b^2=a+b\\ a^2+b^2=1\\ a,b\in [0;1]\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (a+b)(a-b-1)=0\\ a^2+b^2=1\\ a,b\in [0;1]\end{matrix}\right.\)
Nếu $a+b=0$. Vì $a,b\geq 0$ nên $a=b=0$ (điều này vô lý vì $a^2+b^2=1$)
Nếu $a-b-1=0$
$\Leftrightarrow a=b+1$. Do $a\leq 1$ và $b\geq 0$ nên điều này xảy ra khi $b=0; a=1$
$\Rightarrow x=2n\pi \pm \frac{\pi}{2}$ với $n$ nguyên.