Question

Câu 1: Giải các phương trình sau:

a, \(\sqrt{2}sinx-cosx=\sqrt{2}\)

b, sin7x+ \(\sqrt{3}\) cos7x =\(\sqrt{2}\)

c, 5cos2x-12sinx=13

d, sinx+cosx=\(\sqrt{2}\)

e, \(\frac{1+\sqrt{3}}{2\sqrt{2}}\)cosx+ \(\frac{1-\sqrt{3}}{2\sqrt{2}}\)sinx= \(\frac{1}{2}\)

Câu 2: giải các phương trình sau:

a, \(\sqrt{3}\)tanx-6cotx+2\(\sqrt{3}\) - 3=0

b, \(\frac{1-sin2x}{2sinx}\)=sinx

c, \(\sqrt{3}sinx-cosx=1\)

d, \(2sin3x+\sqrt{5}cos3x=3\)

e, sinx(cosx+2sinx)+1=cos²x-2

Answer 1

a.

\(\Leftrightarrow\frac{\sqrt{2}}{\sqrt{3}}sinx-\frac{1}{\sqrt{3}}cosx=\frac{\sqrt{2}}{\sqrt{3}}\)

Đặt \(\frac{\sqrt{2}}{\sqrt{3}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sinx.cosa-cosx.sina=cosa\)

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

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

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

Answer 2

b.

\(\frac{1}{2}sin7x+\frac{\sqrt{3}}{2}cos7x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\7x+\frac{\pi}{3}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{84}+\frac{k2\pi}{7}\\x=\frac{5\pi}{84}+\frac{k2\pi}{7}\end{matrix}\right.\)

c.

\(\Leftrightarrow\frac{5}{13}cos2x-\frac{12}{13}sin2x=1\)

Đặt \(\frac{5}{13}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cos2x.cosa-sin2x.sina=1\)

\(\Leftrightarrow cos\left(2x+a\right)=1\)

\(\Leftrightarrow2x+a=k2\pi\)

\(\Leftrightarrow x=-\frac{a}{2}+k\pi\)

Answer 3

d.

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

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

\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)

e.

\(\Leftrightarrow cosx.cos\left(\frac{\pi}{12}\right)-sinx.sin\left(\frac{\pi}{12}\right)=\frac{1}{2}\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{12}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{12}=\frac{\pi}{3}+k2\pi\\x+\frac{\pi}{12}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 4

2.a.

ĐKXĐ: ...

\(\sqrt{3}tanx-\frac{6}{tanx}+2\sqrt{3}-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-2\\tanx=\sqrt{3}\end{matrix}\right.\)

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

b.

ĐKXĐ: \(x\ne k\pi\)

\(1-sin2x=2sin^2x\)

\(\Leftrightarrow1-2sin^2x-sin2x=0\)

\(\Leftrightarrow cos2x-sin2x=0\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow...\)

Answer 5

c.

\(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

d.

\(\frac{2}{3}sin3x+\frac{\sqrt{5}}{3}cos3x=1\)

Đặt \(\frac{2}{3}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sin3x.cosa+cos3x.sina=1\)

\(\Leftrightarrow sin\left(3x+a\right)=1\)

\(\Leftrightarrow3x+a=\frac{\pi}{2}+k2\pi\)

Answer 6

Câu 1:

b. \(sin7x+\sqrt{3}cos7x=\sqrt{2}\)

\(\Leftrightarrow\frac{1}{2}sin7x+\frac{\sqrt{3}}{2}cos7x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow cos\frac{\pi}{3}.sin7x+sin\frac{\pi}{3}.cos7x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi,k\in Z\\7x+\frac{\pi}{3}=\frac{3}{4}\pi+k2\pi,k\in Z\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{84}+k\frac{2}{7}\pi\\x=\frac{5\pi}{84}+k\frac{2}{7}\pi\end{matrix}\right.k\in Z}\)

Vậy....

Answer 7

e.

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

\(\Leftrightarrow\frac{1}{2}sin2x+1-cos2x+3-\frac{1}{2}\left(1+cos2x\right)=0\)

\(\Leftrightarrow sin2x-3cos2x=-7\)

Do \(1^2+\left(-3\right)^2< \left(-7\right)^2\) nên pt đã cho vô nghiệm

Answer 8

c. \(5cos2x-12sinx=13\)

\(\Leftrightarrow5-10sin^2x-12sinx-13=0\)

\(\Leftrightarrow-10sin^2x-12sinx-8=0\)

Phương trình vô nghiệm

Vậy.....

d. \(sinx+cosx=\sqrt{2}\)

\(\Leftrightarrow\frac{1}{\sqrt{2}}.sinx+\frac{1}{\sqrt{2}}cosx=1\)

\(\Leftrightarrow cos\frac{\pi}{4}.sinx+sin\frac{\pi}{4}.cosx=1\)

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

\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi,k\in Z\)

\(\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)

Vậy....