Question

Giaỉ các phương trình lượng giác sau:

1. 2sin2x+3sinx=3cosx

2. sin2x-4(sinx-cosx)=4

3. (1+sinx)(1+cosx)=2

4. 2(sinx-cosx)-sin2x-1=0

5. sinx-cosx+4sinxcosx+1=0

6. sinx=2cos\(^3\)x

7. cosx=2sin\(^3\)x

8. 2cos\(^3\)x=sin3x

Answer 1

1.

\(\Leftrightarrow4sinx.cosx+3\left(sinx-cosx\right)=0\)

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(2\left(1-t^2\right)+3t=0\)

\(\Leftrightarrow-2t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=2\left(l\right)\\t=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow sinx-cosx=-\frac{1}{2}\)

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

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

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

Answer 2

2.

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(1-t^2-4t=4\)

\(\Leftrightarrow t^2+4t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=-1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

Answer 3

3.

\(\Leftrightarrow1+cosx+sinx+sinx.cosx=2\)

\(\Leftrightarrow2\left(sinx+cosx\right)+2sinx.cosx-2=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(2t+t^2-1-2=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow sinx+cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\)

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

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

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

Answer 4

4.

Đặt \(sinx-cosx=t\Rightarrow...\)

Phương trình trở thành:

\(2t-\left(1-t^2\right)-1=0\)

\(\Leftrightarrow t^2+2t-2=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

Answer 5

5.

Đặt \(sinx-cosx=t\Rightarrow...\)

Pt trở thành:

\(t+2\left(1-t^2\right)+1=0\)

\(\Leftrightarrow-2t^2+t+3=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=\frac{3}{2}>\sqrt{2}\left(l\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

Answer 6

6.

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

\(\Leftrightarrow\frac{sinx}{cosx}.\frac{1}{cos^2x}=2\)

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

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

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

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

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

Answer 7

7.

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

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

\(\Leftrightarrow1+tan^2x=2tan^3x\)

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

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

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

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

Answer 8

8.

\(\Leftrightarrow2cos^3x=3sinx-4sin^3x\)

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

\(\Leftrightarrow2=\frac{3sinx}{cosx}.\frac{1}{cos^2x}-4tan^3x\)

\(\Leftrightarrow2=3tanx\left(1+tan^2x\right)-4tan^3x\)

\(\Leftrightarrow2=3tanx-tan^3x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)