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a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)
\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)
\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)
(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)
\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)
b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx+\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)
\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)
\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)
(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)
\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)
\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)
- Đặt \(\left\{{}\begin{matrix}\sin^2\dfrac{x}{2}=a\\\sin x+3=b\end{matrix}\right.\)
\(PTTT:a^2-ab+b-1=0\)
\(\Leftrightarrow-b\left(a-1\right)+\left(a-1\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+1-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a-b=-1\end{matrix}\right.\)
- Thay lại vào phương trình ta được :\(\left[{}\begin{matrix}\sin^2\dfrac{x}{2}=1\\\sin^2\dfrac{x}{2}-\sin x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin^2\dfrac{x}{2}=1\\\dfrac{1-\cos x}{2}-\sin x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin^2\dfrac{x}{2}=1\\\cos x+2\sin x=-3\end{matrix}\right.\)
Thấy : \(-\sqrt{5}\le2\sin x+\cos x\le\sqrt{5}\)
\(\Rightarrow2\sin x+\cos x=-3\left(L\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin\dfrac{x}{2}=1\\\sin\dfrac{x}{2}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\\\dfrac{x}{2}=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k4\pi\\x=-\pi+k4\pi\end{matrix}\right.\)\(\left(K\in Z\right)\)
Vậy ....
\(1+\sin\dfrac{x}{2}\sin x-\cos\dfrac{x}{2}\sin^2x=2\cos^2\left(\dfrac{\Pi}{4}-\dfrac{x}{2}\right)\)
\(\Leftrightarrow1+\sin\dfrac{x}{2}\sin x-\cos\dfrac{x}{2}\sin^2x=2\left(\dfrac{\sqrt{2}}{2}\cos\dfrac{x}{2}+\dfrac{\sqrt{2}}{2}\sin\dfrac{x}{2}\right)^2\)
\(\Leftrightarrow1+2\sin^2\dfrac{x}{2}\cos\dfrac{x}{2}-\cos\dfrac{x}{2}\left(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\right)^2=1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\)
\(\Leftrightarrow2\sin^2\dfrac{x}{2}\cos\dfrac{x}{2}-4\cos^3\dfrac{x}{2}\sin^2\dfrac{x}{2}-2\sin\dfrac{x}{2}\cos\dfrac{x}{2}=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\left(\sin\dfrac{x}{2}-2\sin\dfrac{x}{2}\cos^2\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\left(\sin\dfrac{x}{2}-2\sin\dfrac{x}{2}\left(1-\sin^2\dfrac{x}{2}\right)-1\right)=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}.\left(\sin\dfrac{x}{2}-1\right)\left(2\sin^2\dfrac{x}{2}+2\sin\dfrac{x}{2}+1\right)=0\)
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= sin(-600) nên phương trình đã cho tương đương với.png)
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a) Ta có:
sin(x+1)=23⇔[x+1=arcsin23+k2πx+1=π−arcsin23+k2π⇔[x=−1+arcsin23+k2πx=−1+π−arcsin23+k2π;k∈Zsin(x+1)=23⇔[x+1=arcsin23+k2πx+1=π−arcsin23+k2π⇔[x=−1+arcsin23+k2πx=−1+π−arcsin23+k2π;k∈Z
b) Ta có:
sin22x=12⇔1−cos4x2=12⇔cos4x=0⇔4x=π2+kπ⇔x=π8+kπ4,k∈Zsin22x=12⇔1−cos4x2=12⇔cos4x=0⇔4x=π2+kπ⇔x=π8+kπ4,k∈Z
c) Ta có:
cot2x2=13⇔⎡⎢⎣cotx2=√33(1)cotx2=−√33(2)(1)⇔cotx2=cotπ3⇔x2=π3+kπ⇔x=2π3+k2π,k∈z(2)⇔cotx2=cot(−π3)⇔x2=−π3+kπ⇔x=−2π3+k2π;k∈Zcot2x2=13⇔[cotx2=33(1)cotx2=−33(2)(1)⇔cotx2=cotπ3⇔x2=π3+kπ⇔x=2π3+k2π,k∈z(2)⇔cotx2=cot(−π3)⇔x2=−π3+kπ⇔x=−2π3+k2π;k∈Z
d) Ta có:
tan(π12+12x)=−√3⇔tan(π12+12π)=tan(−π3)⇔π12+12=−π3+kπ⇔x=−5π144+kπ12,k∈Z
Vậy nghiệm của phương trình đã cho là: x=−5π144+kπ12,k∈Z
a)
\(sin\left(x+1\right)=\dfrac{2}{3}\Leftrightarrow\left[{}\begin{matrix}x+1=arcsin\dfrac{2}{3}+k2\pi\\x+1=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\dfrac{2}{3}-1+k2\pi\\x=\pi-arcsin\dfrac{2}{3}-1+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\).
Đặt \(t=\dfrac{3\pi}{10}-\dfrac{x}{2}\)\(\Rightarrow\pi-3t=\dfrac{\pi}{10}+\dfrac{3\pi}{2}\)
\(pt\Leftrightarrow2sint=sin\left(\pi-3t\right)\)
\(\Leftrightarrow2sint=3sint-4sin^3t\)
\(\Leftrightarrow sint\left(1-4sin^2t\right)=0\)
\(\Leftrightarrow sint\left(2cos2t\right)=0\)
dễ nhé :3