a) <=> 4sinxcosx -(2cos²x-1)=7sinx+2cosx-4

<=> 2cos²x+(2-4sinx)cosx+7sinx-5=0

- sinx=1 => 2cos²x-2cosx+2=0 

pt trên vn

b) <=> 2sinxcosx-1+2sin²x+3sinx-cosx-1=0

<=> cos(2sinx-1)+2sin²x+3sinx-2=0

<=> cosx(2sinx-1)+(2sinx-1)(sinx+2)=0

<=> (2sinx-1)(cosx+sinx+2)=0

<=> sinx=1/2 hoặc cosx+sinx=-2(vn)

<=> x= \(\frac{\pi}{6}+k2\pi\) hoặc \(x=\frac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

a,Pt \(\Leftrightarrow cosx-sinx=\dfrac{1}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x=-\dfrac{\pi}{4}-arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\) ,\(k\in Z\)

b) Pt \(\Leftrightarrow\dfrac{4}{5}cosx-\dfrac{3}{5}sinx=\dfrac{3}{5}\)

Đặt \(cosa=\dfrac{4}{5}\Rightarrow sina=\dfrac{3}{5}\)

Pttt:\(cosx.cosa-sina.sinx=\dfrac{3}{5}\)

\(\Leftrightarrow cos\left(x+a\right)=\dfrac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-a+arc.cos\left(\dfrac{3}{5}\right)+2k\pi\\x=-a-arc.cos\left(\dfrac{3}{5}\right)+2k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

c) Pt\(\Leftrightarrow\dfrac{3}{5}cos3x+\dfrac{4}{5}.sin3x=1\)

Đặt \(cosa=\dfrac{3}{5}\Rightarrow sina=\dfrac{4}{5}\)

Pttt:\(cos3x.cosa+sin3a.sina=1\)

\(\Leftrightarrow cos\left(3x-a\right)=1\)

\(\Leftrightarrow x=\dfrac{a}{3}+\dfrac{k2\pi}{3}\)(\(k\in Z\))

Vậy...

1)\(1+2sinx=2cosx\)

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

\(\Leftrightarrow\left(cosx-sinx\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow cosx^2+sinx^2-2cosxsinx=\dfrac{1}{4}\)

\(\Leftrightarrow1-2cosxsinx=\dfrac{1}{4}\)

\(\Leftrightarrow2cosxsinx=\dfrac{3}{4}\)

\(\Leftrightarrow sin2x=\dfrac{3}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x=arcsin\dfrac{3}{8}+k\pi\\x=\pi-arcsin\dfrac{3}{8}+k\pi\end{matrix}\right.\) \(\left(K\in Z\right)\)

b) \(4cosx-3sinx=3\)

\(\Leftrightarrow\dfrac{4}{5}cosx-\dfrac{3}{5}sinx=\dfrac{3}{5}\)

Đặt \(cosa=\dfrac{3}{5},sina=\dfrac{4}{5}\)

Khi đó:

\(sinacosx-cosasinx=\dfrac{3}{5}\)

\(\Leftrightarrow sin\left(a-x\right)=\dfrac{3}{5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-x=arcsin\dfrac{3}{5}+k2\pi\\a-x=\pi-arcsin\dfrac{3}{5}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=a-arcsin\dfrac{3}{5}+k2\pi\\x=a-\pi-arcsin\dfrac{3}{5}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

3)\(3cos3x+4sin3x=5\)

\(\Leftrightarrow\dfrac{3}{5}cos3x+\dfrac{4}{5}sin3x=1\)

Đặt \(sina=\dfrac{3}{5},cosa=\dfrac{4}{5}\)

khi đó: \(sinacos3x+cosasin3x=1\)

\(\Leftrightarrow sin\left(a+3x\right)=\dfrac{\pi}{2}\)

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

\(\Leftrightarrow x=\dfrac{\pi}{6}-\dfrac{1}{3}a+k\dfrac{2}{3}\pi\),\(k\in Z\)

Chúc bạn học tốt^^

Đáp án D

Đặt t = 3sin x - 4cos x => -5 ≤ t ≤ 5 (dùng bất đẳng thức bunhiacopxki)

Ta có: y = (3sin x – 4cos x)² – 6sin x + 8cos x

              =      t² – 2t = (t – 2)² -1

Do -5 ≤ t ≤ 5 => 0 ≤ (t – 2)² ≤ 36 => min y = -1

Suy ra yêu cầu bài toán -1 ≥ 2m - 1 ⇔ m ≤ 0.

Xét hàm số  y= ( 3sinx – 4cosx )² – 6sinx + 8cosx

Đáp án B

Chọn A

ĐKXĐ; ...

\(\Leftrightarrow\frac{3}{5}sinx+\frac{4}{5}cosx-1=\frac{1}{5}\left(4tanx-3\right)^2\)

\(\Leftrightarrow sin\left(x+a\right)-1=\frac{1}{5}\left(4tanx-3\right)^2\)

(Trong đó \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{3}{5}\))

Do \(\left\{{}\begin{matrix}sin\left(x+a\right)-1\le0\\\left(4tanx-3\right)^2\ge0\end{matrix}\right.\) \(\forall a;x\) nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(x+a\right)=1\\4tanx-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3sinx+4cosx=5\\4sinx-3cosx=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}sinx=\frac{3}{5}\\cosx=\frac{4}{5}\end{matrix}\right.\) \(\Rightarrow x=arcsin\left(\frac{3}{5}\right)+k2\pi\)

Đáp án C