a, \(cos^2x-cosx=0\)

\(\Leftrightarrow cosx\left(cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)

a, \(cos^2x-cosx=0\)

\(\Leftrightarrow cosx\left(cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)

Vậy...

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)

\(\Leftrightarrow8sin^2x-2sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

Vậy...

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\) 

\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)

\(\Leftrightarrow cos2x=\dfrac{1}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)

Vậy...

e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))

\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)

\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)

Vậy...

\(2sin^2x-3sinx+1=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))  (I)

Có \(0\le x< \dfrac{\pi}{2}\)\(\Leftrightarrow\left[{}\begin{matrix}0\le\dfrac{\pi}{2}+k2\pi< \dfrac{\pi}{2}\\0\le\dfrac{\pi}{6}+k2\pi< \dfrac{\pi}{2}\\0\le\dfrac{5\pi}{6}+k2\pi< \dfrac{\pi}{2}\end{matrix}\right.\)(\(k\in Z\))

\(\Leftrightarrow\left[{}\begin{matrix}0\le\dfrac{1}{2}+2k< \dfrac{1}{2}\\0\le\dfrac{1}{6}+2k< \dfrac{1}{2}\\0\le\dfrac{5}{6}+2k< \dfrac{1}{2}\end{matrix}\right.\) (\(k\in Z\)) \(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{4}\le k< 0\left(1\right)\\-\dfrac{1}{12}\le k< \dfrac{1}{6}\left(2\right)\\-\dfrac{5}{12}\le k< -\dfrac{1}{6}\left(3\right)\end{matrix}\right.\)(\(k\in Z\))

Do k nguyên, từ (1) và (3) \(\Rightarrow k\in\varnothing\)

Từ (2)\(\Rightarrow k=0\)\(\Rightarrow x=\dfrac{\pi}{6}+0.2\pi=\dfrac{\pi}{6}\)

Ý C 

(Hoặc sau khi bạn làm đến đoạn số (I),bạn vẽ đường tròn lượng giác ra sẽ tìm được x)

`5sinx(sinx-1)-cos^2x=3`
`<=>5sin^2x-5sinx-(1-sin^2x)-3=0`
`<=>6sin^2x-5sinx-2=0`

`<=>` \(\left[{}\begin{matrix}sinx=\dfrac{5+\sqrt{73}}{12}\left(L\right)\\sinx=\dfrac{5-\sqrt{73}}{12}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=arcsin\left(\dfrac{5-\sqrt{73}}{12}\right)+k2\pi\\x=\pi-arcsin\left(\dfrac{5-\sqrt{73}}{12}\right)+k2\pi\end{matrix}\right.\)

Vậy PT có 2 họ nghiệm như trên.

1.

\(cos^2x-5sin^2x-5cosx+4=0\)

\(\Leftrightarrow cos^2x+5-5sin^2x-5cosx-1=0\)

\(\Leftrightarrow6cos^2x-5cosx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{1}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm arccos\left(-\dfrac{1}{6}\right)+k2\pi\end{matrix}\right.\)

2. 

ĐK: \(x\ne k\pi\)

\(\dfrac{\sqrt{3}}{sin^2x}=cotx+\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(\dfrac{1}{sin^2x}-1\right)-cotx=0\)

\(\Leftrightarrow\sqrt{3}cot^2x-cotx=0\)

\(\Leftrightarrow cotx\left(\sqrt{3}cotx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=0\\cotx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=0\\cotx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

1.

\(4cos^2x-4\sqrt{3}cosx+3=0\)

\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)^2=0\)

\(\Leftrightarrow cosx=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k2\pi\)

2.

\(2sin^2x-7sinx+3=0\)

\(\Leftrightarrow\left(sinx-3\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=3\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

Với \(sinx=3\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(3\right)+k2\pi\\x=\pi-arcsin\left(3\right)+k2\pi\end{matrix}\right.\)

Với \(sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

`2sin^2x-cosx+7/2=0`

`<=>2(1-cos^2x)-cosx+7/2=0`

`<=>-2cos^2x-cosx+11/2=0`

`<=>` \(\left[{}\begin{matrix}cosx=\dfrac{-1+3\sqrt{5}}{4}\left(L\right)\\cosx=\dfrac{-1-3\sqrt{5}}{4}\left(L\right)\end{matrix}\right.\)

Vậy PTVN.

Gọi \(x_0\) là hoành độ của điểm thuộc (C) mà tại đó tiếp xuyến của (C) song song với (d)

Ta có \(y'\left(x_0\right)=\frac{3}{\left(x_0+2\right)^2}\), hệ số góc của \(\left(d\right):3x-y+15=0\) là 3

Suy ra \(\frac{3}{\left(x_0+2\right)^2}=3\Leftrightarrow\orbr{\begin{cases}x_0+2=1\\x_0+2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x_0=-1\\x_0=-3\end{cases}}\)

Hai điểm cần tìm là \(A\left(-1;-2\right),B\left(-3;4\right)\). Vậy \(S=-2+4=2.\)

Bạn có thể tìm 1 số tài liệu trên mạng nha .

Tham khảo :

Ta có : \(x\in\left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right]\)

\(\Rightarrow\cos x\ge0\)

Mà \(-1\le\cos x\le1\)

\(\Rightarrow0\le\cos x\le1\)

\(\Leftrightarrow2\le3\cos x+2\le5\)

Vậy \(\left\{{}\begin{matrix}Min=2\left(x=\pm\dfrac{\pi}{2}+k\pi\right)\\Max=5\left(x=k2\pi\right)\end{matrix}\right.\) \(\forall x\in\left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right]\)

nhìn cái bài mà mệt

f) \(F=sin5^o.sin15^o.sin25^o...sin75^o.sin85^o\\ =\left(sin5^o.sin85^o\right)\left(sin15^o.sin75^o\right)\left(sin25^o.sin65^o\right)\left(sin35^o.sin55^0\right).sin45^o\)

\(=-\dfrac{1}{2}\left(cos90^o-cos80^o\right).-\dfrac{1}{2}\left(cos90^o-cos60^o\right).-\dfrac{1}{2}\left(cos90^o-cos40^o\right).-\dfrac{1}{2}\left(cos90^o-cos20^o\right).sin45^o\)

\(=\dfrac{1}{16}\left(-cos80^o\right).\left(-cos60^o\right)\left(-cos40^o\right)\left(-cos20^o\right)\dfrac{\sqrt{2}}{2}\)

\(\dfrac{\sqrt{2}}{32}.cos20^o.cos40^o.cos60^o.cos80^o\)

nhân cả 2 vế với sin20o

\(F.sin20^o=\dfrac{\sqrt{2}}{2}.sin20^o.cos20^o.cos40^o.cos80^o.cos60^o\)

\(=\dfrac{\sqrt{a}}{64}.sin40^o.cos40^o.cos80^o.cos60^o\)

\(\dfrac{\sqrt{2}}{128}sin80^o.cos80^o.cos60^o\\ =\dfrac{\sqrt{2}}{256}.sin160^o.\dfrac{1}{2}\\ =\dfrac{\sqrt{2}}{512}sin160^o\\ =\dfrac{\sqrt{2}}{512}.sin\left(180^o-20^o\right)\\ =\dfrac{\sqrt{2}}{512}sin20^o\Rightarrow F=\dfrac{\sqrt{2}}{512}\)

bấm cái này mệt lắm G= 3/256, E= 1/32,