Tìm hiệu của số tròn chục lớn nhất có 2 chữ số 

P=​(1−cos2x)2+6cos2x+3cos4x​+(1−sin2x)2+6sin2x+3sin4x​=4cos4x+4cos2x+1​+4sin4x+4sin2x+1​=(2cos2x+1)2​+(2sin2x+1)2​=2cos2x+1+2sin2x+1=3​

Vậy $P$ không phụ thuộc vào $x$.

1.

Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng

2.a

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)

\(\Leftrightarrow1+tan^2x=4tanx+6\)

\(\Leftrightarrow tan^2x-4tanx-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)

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

2b.

Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)

\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow sin^3t=sint+cost\)

\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)

\(\Leftrightarrow sint.cos^2t+cost=0\)

\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

2c.

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{4}\)

\(\dfrac{4sinx.cos2x}{sin4x}+\dfrac{2cos2x}{sin4x}=\dfrac{2}{sin4x}\)

\(\Leftrightarrow2sinx.cos2x+cos2x=1\)

\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)

\(\Leftrightarrow sinx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\cos2x-sinx=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)

a.

Tìm min:

$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$

Vậy $y_{\min}=2$

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Mặt khác: 

$y=4\sin x(\sin x+1)-8(\sin x+1)+11$

$=(\sin x+1)(4\sin x-8)+11$

$=4(\sin x+1)(\sin x-2)+11$

Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$

$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$

$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$

Vậy $y_{\max}=11$

b.

$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$

$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$

Vậy $y_{\max}=4$.

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Mặt khác:

$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$

$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$

$=(1+\sin x)(3-\sin x)$

Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$

$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$

Vậy $y_{\min}=0$

c.

$y=\sin ^4x-2\cos ^2x+1=\sin ^4x-2(1-\sin ^2x)+1$

$=\sin ^4x+2\sin ^2x-1$

$=(\sin ^4x-1)+(2\sin ^2x-2)+2$

$=(\sin ^2x-1)(\sin ^2x+1)+2(\sin ^2x-1)+2$

$=(\sin ^2x-1)(\sin ^2x+3)+2$

Vì $\sin x\in [-1;1]$ nên $\sin ^2x\leq 1$

$\Rightarrow (\sin ^2x-1)(\sin ^2x+3)\leq 0$

$\Rightarrow y=(\sin ^2x-1)(\sin ^2x+3)+2\leq 2$

Vậy $y_{\max}=2$

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$y=\sin ^4x+2\sin ^2x-1=\sin ^2x(\sin ^2x+2)-1$

Vì $\sin ^2x\geq 0$ nên $\sin ^2x(\sin ^2x+2)\geq 0$

$\Rightarrow y=\sin ^2x(\sin ^2x+2)-1\geq 0-1=-1$
Vậy $y_{\min}=-1$

a) Dễ thấy cosx = 0 không thỏa mãn phương trình đã cho nên chiaw phương trình cho cos²x ta được phương trình tương đương 2tan²x + tanx - 3 = 0.

Đặt t = tanx thì phương trình này trở thành

2t² + t - 3 = 0 ⇔ t ∈ {1 ; }.

Vậy

b) Thay 2 = 2(sin²x + cos²x), phương trình đã cho trở thành

3sin²x - 4sinxcosx + 5cos²x = 2sin²x + 2cos²x

⇔ sin²x - 4sinxcosx + 3cos²x = 0

⇔ tan²x - 4tanx + 3 = 0

⇔

⇔ x = + kπ ; x = arctan3 + kπ, k ∈ Z.

c) Thay sin2x = 2sinxcosx ; = (sin²x + cos²x) vào phương trình đã cho và rút gọn ta được phương trình tương đương

sin²x + 2sinxcosx - cos²x = 0 ⇔ tan²x + 4tanx - 5 = 0 ⇔

⇔ x = + kπ ; x = arctan(-5) + kπ, k ∈ Z.

d) 2cos²x - 3√3sin2x - 4sin²x = -4

⇔ 2cos²x - 3√3sin2x + 4 - 4sin²x = 0

⇔ 6cos²x - 6√3sinxcosx = 0 ⇔ cosx(cosx - √3sinx) = 0

⇔

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

6.

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

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

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

1.

\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)

\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)

\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)

\(\Leftrightarrow4cos^32x=\sqrt{2}\)

\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

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

ghi đề rõ xíu đi

\(\Leftrightarrow cos^4x+sin^4x+\dfrac{1}{2}\left[sin\left(3x-\dfrac{pi}{4}+x-\dfrac{pi}{4}\right)+sin\left(3x-\dfrac{pi}{4}-x+\dfrac{pi}{4}\right)\right]-\dfrac{3}{2}=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[sin\left(4x-\dfrac{pi}{2}\right)+sin2x\right]-\dfrac{3}{2}=0\)

=>\(-\dfrac{1}{2}sin^22x-\dfrac{1}{2}+\dfrac{1}{2}\left[-sin\left(\dfrac{pi}{2}-4x\right)+sin2x\right]=0\)

=>\(-sin^22x-1-cos4x+sin2x=0\)

=>\(-sin^22x-1-\left(1-2sin^22x\right)+sin2x=0\)

=>\(-sin^22x-1-1+2sin^22x+sin2x=0\)

=>\(sin^22x+sin2x-2=0\)

=>sin2x-1=0

=>sin2x=1

=>2x=pi/2+k2pi

=>x=pi/4+kpi

a.

\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

c.

\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)

\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)

\(\Leftrightarrow...\)