Với \(cosx=0\) không phải nghiệm

Với \(cosx\ne0\) , chia 2 vế cho \(cos^3x\) ta được:

\(tan^3x-5tan^2x-3tanx+3=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-6tanx+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3-\sqrt{6}\\tanx=3+\sqrt{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(3-\sqrt{6}\right)+k\pi\\x=arctan\left(3+\sqrt{6}\right)+k\pi\end{matrix}\right.\)

\(a\text{) }sin^3x+cos^3x=sinx+cosx\\ \Leftrightarrow\left(sinx+cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)=sinx+cosx\\ \Leftrightarrow-\frac{1}{2}sin2x\left(sinx+cosx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=-cosx=sin\left(x-\frac{\pi}{2}\right)\\sin2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{2}-x+a2\pi\\2x=b\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{4}+a\pi\\x=\frac{b\pi}{2}\end{matrix}\right.\)

\(\text{b) }sin^3x+2sin^2x\cdot cosx-3cos^3x=0\\ \Leftrightarrow\left(sin^3x-cos^3x\right)+2cosx\cdot\left(sin^2x-cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(sinx\cdot cosx+1\right)+\left(sinx-cosx\right)\left(2sinx\cdot cosx+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(3sinx\cdot cosx+1+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(\frac{3}{2}sin2x+2+cos2x\right)=0\)

Với \(sinx-cosx=0\)

\(\Leftrightarrow sinx=cosx=sin\left(\frac{\pi}{2}-x\right)\\ \Leftrightarrow x=\frac{\pi}{2}-x+a2\pi\\ \Leftrightarrow x=\frac{\pi}{4}+a\pi\)

Với \(\frac{3}{2}sin2x+2+cos2x=0\)

\(\Leftrightarrow sin^22x+\left(\frac{3}{2}sin2x+2\right)^2=1\left(VN\right)\)

\(\text{c) }3cos^4x-4cos^2x\cdot sin^2x-sin^4x=0\)

Nhận thấy sinx=0 không là nghiệm pt.

Chia cả 2 vế cho sin⁴x ta được

\(pt\Leftrightarrow\frac{3cos^4x}{sin^4x}-\frac{4cos^2x}{sin^2x}-1=0\\ \Leftrightarrow3cot^4x-4cot^2x-1=0\\ \Leftrightarrow cot^2x=\frac{2+\sqrt{7}}{3}\\ \Leftrightarrow cotx=\pm\sqrt{\frac{2+\sqrt{7}}{3}}\\ \Leftrightarrow x=arccot\left(\pm\sqrt{\frac{2+\sqrt{7}}{3}}\right)+k2\pi\)

d) kiểm tra đề.

a/

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(3tan^3x+2tan^2x=tanx\)

\(\Leftrightarrow tanx\left(3tan^2x+2tanx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=0\\3tan^2x+2tanx-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=0\\tanx=-1\\tanx=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{4}+k\pi\\x=arctan\left(\frac{1}{3}\right)+k\pi\end{matrix}\right.\)

b/ \(\Leftrightarrow3sinx+cos^3x=5sinx.cos^2x\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(3tanx.\frac{1}{cos^2x}+1=5tanx\)

\(\Leftrightarrow3tanx\left(1+tan^2x\right)-5tanx+1=0\)

\(\Leftrightarrow3tan^3x-2tanx+1=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(3tan^2x-3tanx+1\right)=0\)

\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)

a/ Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow2tan^3x=\frac{1}{cos^2x}\)

\(\Leftrightarrow2tan^3x=1+tan^2x\)

\(\Leftrightarrow2tan^3x-tan^2x-1=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(2tan^2x+tanx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\2tan^2x+tanx+1=0\left(vn\right)\end{matrix}\right.\)

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

b/

\(\Leftrightarrow sin^3x-sinx+cos^3x+cosx=0\)

\(\Leftrightarrow-sinx\left(1-sin^2x\right)+cos^3x+cosx=0\)

\(\Leftrightarrow-sinx.cos^2x+cos^3x+cosx=0\)

\(\Leftrightarrow cosx\left(cos^2x-sinx.cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\cos^2x-sinx.cosx+1=0\left(1\right)\end{matrix}\right.\)

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

\(\Leftrightarrow cos2x-sin2x=-3\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=\frac{3}{\sqrt{2}}>1\left(vn\right)\)

c/

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(4+2tan^3x-3tanx.\frac{1}{cos^2x}=0\)

\(\Leftrightarrow2tan^3x-3tanx\left(1+tan^2x\right)+4=0\)

\(\Leftrightarrow-tan^3x-3tanx+4=0\)

\(\Leftrightarrow\left(1-tanx\right)\left(tan^2x+tanx+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x+tanx+4=0\left(vn\right)\end{matrix}\right.\)

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

d/

\(\Leftrightarrow2cos^3x+2sinx-6sin^2x.cosx=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(2+2tanx.\frac{1}{cos^2x}-6tan^2x=0\)

\(\Leftrightarrow1+tanx\left(1+tan^2x\right)-3tan^2x=0\)

\(\Leftrightarrow tan^3x-3tan^2x+tanx+1=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x-2tanx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x-2tanx-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=1-\sqrt{2}\\tanx=1+\sqrt{2}\end{matrix}\right.\)

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

Với \(cosx=0\) không phải nghiệm

Với \(cosx\ne0\) , chia 2 vế cho \(cos^3x\):

\(4tan^3x+3tan^2x-tanx.\left(1+tan^2x\right)-1=0\)

\(\Leftrightarrow3tan^3x+3tan^2x-tanx-1=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(3tan^2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\dfrac{1}{\sqrt{3}}\\tanx=-\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

d.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^4x\)

\(tan^4x-3tan^2x-4tanx-3=0\)

\(\Leftrightarrow\left(tan^2x+tanx+1\right)\left(tan^2x-tanx-3\right)=0\)

\(\Leftrightarrow tan^2x-tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1-\sqrt{13}}{2}\\tanx=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\frac{1-\sqrt{13}}{2}\right)+k\pi\\x=arctan\left(\frac{1+\sqrt{13}}{2}\right)+k\pi\end{matrix}\right.\)

mọi người giúp hộ mình nhanh với

a.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(2tan^3x+4=3tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+4=3tanx+3tan^3x\)

\(\Leftrightarrow tan^3x+3tanx-4=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x+tanx+4\right)=0\)

\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)