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\(\Leftrightarrow tan^2x+cot^2x=2\left(cos^4x+sin^4x+2sin^2x.cos^2x\right)\)
\(\Leftrightarrow tan^2x+cot^2x=2\left(sin^2x+cos^2x\right)^2\)
\(\Leftrightarrow tan^2x+cot^2x=2\)
\(\Leftrightarrow\left(tanx-cotx\right)^2=0\)
\(\Leftrightarrow tanx=cotx=tan\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow x=\frac{\pi}{2}-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
c.
\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)
\(\Leftrightarrow2cos\left(x+12^0\right)=1\)
\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)
2.
Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:
\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)
\(\Rightarrow-1\le m\le\dfrac{1}{2}\)
a.
\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
b.
\(2x-10^0=arccot\left(4\right)+k180^0\)
\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)
ĐK: \(x\ne k\pi\)
\(\dfrac{1+sin2x+cos2x}{1+cot^2x}=sinx.\left(sin2x+2sin^2x\right)\)
\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{cos^2x+sin^2x}{sin^2x}}=sinx.\left(2sinx.cosx+2sin^2x\right)\)
\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{1}{sin^2x}}=2sin^2x.\left(cosx+sinx\right)\)
\(\Leftrightarrow1+sin2x+cos2x=2cosx+2sinx\)
\(\Leftrightarrow1+2sinx.cosx+2cos^2x-1=2cosx+2sinx\)
\(\Leftrightarrow\left(cosx-1\right).\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(cosx-1\right).sin\left(x+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x+\dfrac{\pi}{4}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
1.
Đặt \(f\left(x\right)=\left(m^2+1\right)x^3-2m^2x^2-4x+m^2+1\)
\(f\left(x\right)\) xác định và liên tục trên R
\(f\left(x\right)\) có bậc 3 nên có tối đa 3 nghiệm (1)
\(f\left(0\right)=m^2+1>0\) ; \(\forall m\)
\(f\left(1\right)=\left(m^2+1\right)-2m^2-4+m^2+1=-2< 0\) ;\(\forall m\)
\(\Rightarrow f\left(0\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(0;1\right)\) (2)
\(f\left(2\right)=8\left(m^2+1\right)-8m^2-8+m^2+1=m^2+1>0\)
\(\Rightarrow f\left(1\right).f\left(2\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(1;2\right)\) (3)
\(f\left(-3\right)==-27\left(m^2+1\right)-18m^2+12+m^2+1=-44m^2-14< 0\)
\(\Rightarrow f\left(-3\right).f\left(0\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(-3;0\right)\) (4)
Từ (1); (2); (3); (4) \(\Rightarrow f\left(x\right)=0\) có đúng 3 nghiệm phân biệt
2.
Đặt \(t=g\left(x\right)=x.cosx\)
\(g\left(x\right)\) liên tục trên R và có miền giá trị bằng R \(\Rightarrow t\in\left(-\infty;+\infty\right)\)
\(f\left(t\right)=t^3+m\left(t-1\right)\left(t+2\right)\)
Hàm \(f\left(t\right)\) xác định và liên tục trên R
\(f\left(1\right)=1>0\)
\(f\left(-2\right)=-8< 0\)
\(\Rightarrow f\left(1\right).f\left(-2\right)< 0\Rightarrow f\left(t\right)=0\) luôn có ít nhất 1 nghiệm thuộc \(\left(-2;1\right)\)
\(\Rightarrow f\left(x\right)=0\) luôn có nghiệm với mọi m
\(y=\sqrt{\dfrac{\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+m-1}{2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}-5m}}\)
Hàm xác định trên R khi:
TH1: \(\left\{{}\begin{matrix}\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+m-1\ge0\\2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}-5m>0\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow\left\{{}\begin{matrix}-m\le\min\limits_R\left(\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)-1\right)=-1-\sqrt{2}\\5m< \min\limits_R\left(2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}\right)=\dfrac{327}{32}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m\ge1+\sqrt{2}\\m< \dfrac{327}{160}\end{matrix}\right.\) \(\Rightarrow m\in\varnothing\)
Th2: \(\left\{{}\begin{matrix}\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+m-1\le0\\2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}-5m< 0\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow\left\{{}\begin{matrix}m\le\min\limits_R\left(\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)-1\right)=-1-\sqrt{2}\\5m>\max\limits_R\left(2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}\right)=14\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m\le-1-\sqrt{2}\\m>\dfrac{14}{5}\end{matrix}\right.\) \(\Rightarrow m\in\varnothing\)
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\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt[]{3}sin2x-cos2x\)
\(\Leftrightarrow4.\left(-\dfrac{1}{2}\right)\left[sin\left(x+\dfrac{\pi}{3}+x-\dfrac{\pi}{6}\right)+sin\left(x+\dfrac{\pi}{3}-x+\dfrac{\pi}{6}\right)\right]=m^2+2.\left[\dfrac{\sqrt[]{3}}{2}.sin2x-\dfrac{1}{2}.cos2x\right]\)
\(\Leftrightarrow2\left[sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(2x-\dfrac{\pi}{6}\right)\right]=m^2+2\)
\(\Leftrightarrow2.2sin2x.cos\dfrac{\pi}{6}=m^2+2\)
\(\Leftrightarrow2.2sin2x.\dfrac{\sqrt[]{3}}{2}=m^2+2\)
\(\Leftrightarrow2\sqrt[]{3}sin2x.=m^2+2\)
\(\Leftrightarrow sin2x.=\dfrac{m^2+2}{2\sqrt[]{3}}\)
Phương trình có nghiệm khi và chỉ khi
\(\left|\dfrac{m^2+2}{2\sqrt[]{3}}\right|\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m^2+2}{2\sqrt[]{3}}\ge-1\\\dfrac{m^2+2}{2\sqrt[]{3}}\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2\ge-2\left(1+\sqrt[]{3}\right)\left(luôn.đúng\right)\\m^2\le2\left(1-\sqrt[]{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow-\sqrt[]{2\left(1-\sqrt[]{3}\right)}\le m\le\sqrt[]{2\left(1-\sqrt[]{3}\right)}\)
C) Pt \(\Rightarrow m\cdot\dfrac{1-cos2x}{2}-\left(m-1\right)sin2x+\left(2m+1\right)\cdot\dfrac{1+cos2x}{2}=0\)
\(\Rightarrow\left(m+1\right)cos2x-\left(2m-2\right)sin2x=-1-3m\)
Pt có nghiệm: \(\Leftrightarrow\) \(\left(m+1\right)^2+\left[-\left(2m-2\right)\right]^2\ge\left(1+3m\right)^2\)
\(\Rightarrow\dfrac{-3-\sqrt{13}}{2}\le m\le\dfrac{-3+\sqrt{13}}{2}\)
Pt vô nghiệm: \(\Rightarrow\left\{{}\begin{matrix}m>\dfrac{-3+\sqrt{13}}{2}\\m< \dfrac{-3-\sqrt{13}}{2}\end{matrix}\right.\)
c/
\(\Leftrightarrow sin3x-\sqrt{3}cos3x=sinx+\sqrt{3}cosx\)
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=x+\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{3}=\frac{2\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
a/
\(\Leftrightarrow\sqrt{3}cos2x-\left(sin^2x+cos^2x-2sinx.cosx\right)=2\)
\(\Leftrightarrow\sqrt{3}cos2x-1+sin2x=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x=\frac{3}{2}\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=\frac{3}{2}\)
Vế phải lớn hơn 1 nên pt vô nghiệm
b/
\(\Leftrightarrow\frac{5}{2}\left(1+cos2x\right)+2sin2x=4\)
\(\Leftrightarrow4sin2x+5cos2x=3\)
\(\Leftrightarrow\frac{4}{\sqrt{41}}sin2x+\frac{5}{\sqrt{41}}cos2x=\frac{3}{\sqrt{41}}\)
Đặt \(\frac{4}{\sqrt{41}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sin2x.cosa+cos2x.sina=\frac{3}{\sqrt{41}}\)
\(\Leftrightarrow sin\left(2x+a\right)=\frac{3}{\sqrt{41}}=sinb\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+a=b+k2\pi\\2x+a=\pi-b+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{b}{2}-\frac{a}{2}+k\pi\\x=\frac{\pi}{2}-\frac{a}{2}-\frac{b}{2}+k\pi\end{matrix}\right.\)
a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)
b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)
\(\Leftrightarrow2\left(sin^2x+cos^2x\right)-4sin^2x.cos^2x+\left(m+2\right)sin2x-2m-2=0\)
\(\Leftrightarrow-sin^22x+\left(m+2\right)sin2x-2m=0\)
\(\Leftrightarrow-sin^22x+2sin2x+m.sin2x-2m=0\)
\(\Leftrightarrow-sin2x\left(sin2x-2\right)+m\left(sin2x-2\right)=0\)
\(\Leftrightarrow\left(m-sin2x\right)\left(sin2x-2\right)=0\)
\(\Leftrightarrow sin2x=m\)
Mà \(-1\le sin2x\le1\) \(\Rightarrow-1\le m\le1\)