\(cot\left(x+\dfrac{\pi}{4}\right)=-1\)

\(\Rightarrow x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k\pi\)

\(\Rightarrow x=-\dfrac{\pi}{2}+k\pi\)

\(\Rightarrow x=\left\{-\dfrac{\pi}{2};\dfrac{\pi}{2};\dfrac{3\pi}{2};\dfrac{5\pi}{2}\right\}\) có 4 nghiệm

\(cosx=\dfrac{\sqrt{3}}{2}\Rightarrow x=\pm\dfrac{\pi}{6}+k2\pi\)

\(x\in\left[\pi;2\pi\right]\Rightarrow x=\dfrac{11\pi}{6}\)

\(\Rightarrow ab=11.6=66\)

\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)

\(\Leftrightarrow cos2x=1-sinx\)

\(\Leftrightarrow1-2sin^2x=1-sinx\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin^2x.cosx=1\)

\(\Leftrightarrow1-3sin^2x.cos^2x+3sin^2x.cosx=1\)

\(\Leftrightarrow3sinx.cosx\left(1-cosx\right)=0\)

\(\Leftrightarrow sin2x\left(1-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cosx=1\end{matrix}\right.\)

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

\(\Rightarrow x=\dfrac{k\pi}{2}\)

\(cos3x+cos5x-cos4x=0\)

\(\Leftrightarrow2cos4x.cosx-cos4x=0\)

\(\Leftrightarrow cos4x\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(sin2x+sinx+2cos3x.sinx=1-cos\left(4x+\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow sin2x+sinx+sin4x-sin2x=1+sin4x\)

\(\Leftrightarrow sinx=1\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow sin2x+2\cdot\dfrac{1+cos2x}{2}-sinx-cosx=1\)

\(\Leftrightarrow sin2x+cos2x=sinx+cosx\)

\(\Leftrightarrow sin\left(2x+\dfrac{pi}{4}\right)=sin\left(x+\dfrac{pi}{4}\right)\)

=>2x+pi/4=x+pi/4+k2pi hoặc 2x+pi/4=-x+3/4pi+k2pi

=>x=k2pi hoặc x=1/6pi+k2pi/3

`[tan x]/[1+tan^2 x]=2cos 2x cos x+sin x-1-cos 3x`      `ĐK: x \ne \pi/2+k\pi`

`<=>tan x. cos^2 x=cos x+cos 3x+sin x-1-cos 3x`

`<=>sin x cos x=sin x +cos x-1`

Đặt `sin x+cos x=t`     `(t in [-\sqrt{2};\sqrt{2}])`

   `=>sin x cos x=[t^2-1]/2`

Ptr có dạng: `[t^2-1]/2=t-1`

           `<=>t^2-2t+1=0<=>t=1`  (t/m)

  `=>sin x+cos x=1`

`<=>sin (x+\pi/4)=1/\sqrt{2}`

`<=>[(x+\pi/4=\pi/4+k2\pi),(x+\pi/4=[3\pi]/4+k2\pi):}`

`<=>[(x=k2\pi(t//m)),(x=\pi/2+k2\pi(ko t//m)):}`

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(tanx.cos^2x=2cos2x.cosx+sinx-1-cos3x\)

\(\Leftrightarrow sinx.cosx=cos3x+cosx+sinx-1-cos3x\)

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

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

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

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

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

a.

Do \(SP>PC\) nên MP ko song song BC và NP không song song CD

Trong mp (SBC), nối MP kéo dài cắt BC tại E

Trong mp (SCD), nối NP kéo dài cắt CD tại F

\(\Rightarrow\left\{{}\begin{matrix}E\in\left(MNP\right)\cap\left(ABCD\right)\\F\in\left(MNP\right)\cap\left(ABCD\right)\end{matrix}\right.\)

\(\Rightarrow EF=\left(MNP\right)\cap\left(ABCD\right)\)

b.

\(O=AC\cap BD\Rightarrow SO=\left(SAC\right)\cap\left(SBD\right)\)

Trong mp (SBD), nối MN cắt SO tại K

\(\left\{{}\begin{matrix}K\in MN\in\left(MNP\right)\\K\in SO\end{matrix}\right.\) \(\Rightarrow K=SO\cap\left(MNP\right)\) (1)

c.

Do \(\left(CAD\right)\equiv\left(ABCD\right)\) (hai mp có 3 điểm chung là A, C, D)

\(\Rightarrow EF=\left(CAD\right)\cap\left(MNP\right)\)

(Tại sao người ta lại cho câu này khi đã có câu a nhỉ?)

d.

\(K\in SO\in\left(SAC\right)\) (2)

(1);(2) \(\Rightarrow K\in\left(MNP\right)\cap\left(SAC\right)\) (3)

\(\left\{{}\begin{matrix}P\in\left(MNP\right)\\P\in SA\in\left(SAC\right)\end{matrix}\right.\) \(\Rightarrow P\in\left(SAC\right)\cap\left(MNP\right)\) (4)

\(\left\{{}\begin{matrix}I\in EF\in\left(MNP\right)\\I\in AC\in\left(SAC\right)\end{matrix}\right.\) \(\Rightarrow I\in\left(MNP\right)\cap\left(SAC\right)\) (5)

(3);(4);(5) \(\Rightarrow K,I,P\) cùng nằm trên giao tuyến của ((MNP) và (SAC)

Hay K, I, P thẳng hàng

Số cách lấy là:

\(C^3_{14}=364\left(cách\right)\)