a/ \(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\sin\left(\dfrac{\pi}{6}-x\right)=\dfrac{\sqrt{2}}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{6}-x=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{6}-x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)

b/ \(\cos x=0\) ko la nghiem cua pt

\(\cos x\ne0\Rightarrow pt\Leftrightarrow5\tan^2x+\tan x-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\tan x=1\\\tan x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow...\)

a) \(sin^2x+\left(1-\sqrt[]{3}\right)sinxcosx-\sqrt[]{3}cos^2x=0\)

\(\Leftrightarrow tan^2x+\left(1-\sqrt[]{3}\right)tanx-\sqrt[]{3}=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{3\pi}{4}\\tanx=tan\dfrac{\pi}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=tan\dfrac{3\pi}{4}+k\pi\\x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.\left(k\in Z\right)\)

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

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

\(\Leftrightarrow cos2x\left(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot sin2x=\dfrac{-\sqrt{3}-1}{2}\)

\(\Leftrightarrow sin2x\cdot\dfrac{-\sqrt{3}-1}{2}+cos2x\cdot\dfrac{\sqrt{3}-1}{2}=\dfrac{-\sqrt{3}-1}{2}\)

\(\Leftrightarrow sin2x\left(-\sqrt{3}-1\right)+cos2x\left(\sqrt{3}-1\right)=-\sqrt{3}-1\)

\(\Leftrightarrow sin2x\cdot\dfrac{-\sqrt{3}-1}{8}+cos2x\cdot\dfrac{\sqrt{3}-1}{8}=\dfrac{-\sqrt{3}-1}{8}\)

\(\Leftrightarrow sin\left(2x+a\right)=cosa=sin\left(\dfrac{pi}{2}-a\right)\)(với \(cosa=\dfrac{-\sqrt{3}-1}{8}\))

\(\Leftrightarrow\left[{}\begin{matrix}2x+a=\dfrac{pi}{2}-a+k2pi\\2x+a=pi-\dfrac{pi}{2}+a+k2pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-2a+\dfrac{pi}{2}+k2pi\\2x=\dfrac{pi}{2}+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-a+\dfrac{pi}{4}+kpi\\x=\dfrac{pi}{4}+kpi\end{matrix}\right.\)

cái này trên OLM mà

cái này chắc cũng lớp 10 chứ ko thoát đâu

nhưng nếu chỗ =căn 3+1 bỏ căn 3 thì tui ms làm đc

mik lm biếng quá mik chỉ nói cách làm thôi nha bạn

1) chia hai vế cho cos^2(x) \(\sqrt{3}tan^2x+\left(1-\sqrt{3}\right)tanx-1+\left(1-\sqrt{3}\right)\left(1+tan^2x\right)=0\)

đặt t = tanx rr giải thôi =D ( máy 570 thì mode5 3 còn máy 580 thì mode 9 2 2) :)))

2) cx làm cách tương tự chia 2 vế cho cos^2x

3) giữ vế trái bung vế phải ra

\(sin2x-2sin^2x=2-4sin^22x\)

đặt t = sin2x (-1=<t=<1)

4) đẩy sinx cosx qua trái hết

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

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

\(-cos\left(sinxcosx+cos^2x+1\right)=0\)

cái vế đầu cosx=0 bn bik giả rr mà dễ ẹc à còn vế sau thì chia cho cos^2(x) như mấy bài trên rr sau đó đặt t = tanx rr bấm máy là ra thui :))

5)bung cái hằng đẳng thức ra sau đó đặt t=sinx+cosx (t thuộc [-căn(2) ; căn(2)]

khi đó ta có sinxcosx=1/2 sin2x= 1/2t^2 - 1/2

làm đi là ra à

a, \(cos^2x-cosx=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)

a, \(cos^2x-cosx=0\)

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

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

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

Vậy...

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)

\(\Leftrightarrow8sin^2x-2sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

Vậy...

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\) 

\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)

\(\Leftrightarrow cos2x=\dfrac{1}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)

Vậy...

e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))

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

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

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

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow-8sin^2x+2sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\)

Với \(sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Với \(sinx=-\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0\)

\(\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)\)

\(\Leftrightarrow cos^2x=\dfrac{3}{4}\)

\(\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}\)

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

Với \(cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi\)

1/ \(pt\Leftrightarrow\left(3cos^2x-sin^2x\right)\left(cos^2x-sin^2x\right)=0\)

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

\(\Leftrightarrow\left(2cos2x+1\right)cos2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

2/ \(pt\Leftrightarrow\left(sinx-1\right)\left(sin^2x+sinx+6\right)=0\)

\(\Leftrightarrow sinx=1\)

3/ \(pt\Leftrightarrow\dfrac{1-cos2x}{2}-4sin2x+\dfrac{7}{2}\left(1+cos2x\right)=0\)

\(\Leftrightarrow3cos2x-4sin2x=-4\)

\(\Leftrightarrow5\left(\dfrac{3}{5}cos2x-\dfrac{4}{5}sin2x\right)=-4\)

\(\Leftrightarrow cos\left(2x+arccos\dfrac{3}{5}\right)=-\dfrac{4}{5}\)

4,5 giải tương tự câu 3