bài 1 bung công thức sin^6(x) + cos^6(x) là 5/8 + 3/8cos4x = cos4x chuyển vế giải

bài 2 dùng công thức hạ bậc sau đó dùng công thức cộng là ra

a/

\(cos^4x-\left(1-2sin^2x\right)+2sin^6x=0\)

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

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

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

\(\Leftrightarrow sin^2x\left(2sin^4x+sin^2x\right)=0\)

\(\Leftrightarrow sin^4x\left(2sin^2x+1\right)=0\)

\(\Leftrightarrow sinx=0\)

\(\Leftrightarrow x=k\pi\)

b/

\(cos4x=\frac{1}{2}+\frac{1}{2}cos6x\)

\(\Leftrightarrow2\left(2cos^22x-1\right)=1+4cos^32x-3cos2x\)

\(\Leftrightarrow4cos^32x-4cos^22x-3cos2x+3=0\)

\(\Leftrightarrow\left(cos2x-1\right)\left(4cos^22x-3\right)=0\)

\(\Leftrightarrow\left(cos2x-1\right)\left(2cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos4x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)

\(\Rightarrow x=\left\{0;-\frac{11\pi}{12};-\frac{5\pi}{12};\frac{\pi}{12};\frac{7\pi}{12};-\frac{7\pi}{12};-\frac{\pi}{12};\frac{5\pi}{12};\frac{11\pi}{12}\right\}\)

Bạn tự cộng lại

c/

\(\Leftrightarrow2cos^2x-1-\left(2m+1\right)cosx+m+1=0\)

\(\Leftrightarrow2cos^2x-\left(2m+1\right)cosx+m=0\)

\(\Leftrightarrow2cos^2x-cosx-2mcosx+m=0\)

\(\Leftrightarrow cosx\left(2cosx-1\right)-m\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left(cosx-m\right)\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=m\end{matrix}\right.\)

Do \(cosx=\frac{1}{2}\) vô nghiệm trên \(\left(\frac{\pi}{2};\frac{3\pi}{2}\right)\) nên pt có nghiệm khi và chỉ khi \(cosx=m\) có nghiệm trên khoảng đã cho

Mà \(-1< cosx< 0\Rightarrow-1< m< 0\)

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.

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

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

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

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

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

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

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

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

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

\(cos^4x-sin^4x=sin3x+cos4x\)

\(\Leftrightarrow\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=sin3x+cos4x\)

\(\Leftrightarrow cos2x=sin3x+cos4x\)

\(\Leftrightarrow cos4x-cos2x+sin3x=0\)

\(\Leftrightarrow-2sin3x.sinx+sin3x=0\)

\(\Leftrightarrow sin3x\left(1-2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{0;\dfrac{\pi}{3};\dfrac{2\pi}{3};\pi;\dfrac{\pi}{6};\dfrac{5\pi}{6}\right\}\)

\(\Rightarrow\sum x=3\pi\)

a) Pt\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x\left(sin^2x+cos^2x\right)+3sinx.cosx-\dfrac{m}{4}+2=0\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x-\dfrac{3}{2}sin2x-\dfrac{m}{4}+2=0\)

\(\Leftrightarrow-3sin^22x-6sin2x-m+12=0\)

Đặt \(t=sin2x;t\in\left[-1;1\right]\)

Pttt: \(-3t^2-6t-m+12=0\)

\(\Leftrightarrow-3t^2-6t+12=m\) (1)

Đặt \(f\left(t\right)=-3t^2-6t+12;t\in\left[-1;1\right]\) 

Vẽ BBT sẽ tìm được \(f\left(t\right)_{min}=3;f\left(t\right)_{max}=15\)\(\Leftrightarrow3\le f\left(t\right)\le15\)\(\Rightarrow m\in\left[3;15\right]\) thì pt (1) sẽ có nghiệm

mà \(m\in Z\) nên tổng m nguyên để pt có nghiệm là 13 m

Vậy có tổng 13 m nguyên

b) Pt\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(1\right)\\2cos^2x-\left(2m+1\right)cosx+m=0\left(2\right)\end{matrix}\right.\)

Từ (1)\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)

\(x\in\left[0;2\pi\right]\Rightarrow0\le\dfrac{\pi}{2}+k2\pi\le2\pi\)\(\Leftrightarrow-\dfrac{1}{4}\le k\le\dfrac{3}{4}\)\(\Rightarrow k=0\)

Tại k=0\(\Rightarrow x=\dfrac{\pi}{2}\)

Để pt ban đầu có 4 nghiệm pb \(\in\left[0;2\pi\right]\)

\(\Leftrightarrow\) Pt (2) có 3 nghiệm pb khác \(\dfrac{\pi}{2}\)

Xét pt (2) có: \(2cos^2x-\left(2m+1\right)cosx+m=0\)

Vì là phương trình bậc hai ẩn \(cosx\) nên pt (2) chỉ có nhiều nhất ba nghiệm \(\Leftrightarrow\) Pt (2) có một nghiệm cosx=0

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) mà \(x\ne\dfrac{\pi}{2}\)

\(\Rightarrow\) Pt (2) chỉ có nhiều nhất hai nghiệm

\(\Rightarrow\) Pt ban đầu không thể có 4 nghiệm phân biệt

Vậy \(m\in\varnothing\) 

1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)

⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)

⇔ 2cos²x - 5cosx + 2 = 0

⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên

2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)

⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0 

⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)

⇒ sin⁴x + cos⁴x = 48.sin⁴x . cos⁴x

⇔ (sin²x + cos²x)² - 2sin²x. cos²x = 3 . (2sinx.cosx)⁴

⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)² = 3(2sinx.cosx)⁴

⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin⁴2x

⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)

⇔ 1 - 2sin²2x = 0

⇔ cos4x = 0

⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)

⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)

⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)

⇔ sin2x - sin²2x - (1 + cos4x) = 0

⇔ sin2x - sin²2x - 2cos²2x = 0

⇔ sin2x - 2 (cos²2x + sin²2x) + sin²2x = 0

⇔ sin²2x + sin2x - 2 = 0

⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)

⇔ sin2x = 1

⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

4, cos5x + cos2x + 2sin3x . sin2x = 0

⇔ cos5x + cos2x + cosx - cos5x = 0

⇔ cos2x + cosx = 0

⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)

⇔ \(cos\dfrac{3x}{2}=0\)

⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)

⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)

Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)

⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}

Vậy các nghiệm thỏa mãn là các phần tử của tập hợp 

\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)

5, \(\dfrac{cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx}{sin2x-1}=1\)

⇒ \(cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx=sin2x-1\)

⇒ cos²x + 3sin²x + 3\(\sqrt{2}\)sin2x + 1 = 0

⇔ 2 + 2sin²x + 3\(\sqrt{2}\)sin2x = 0

⇔ 2 + 1 - cos2x + 3\(\sqrt{2}\) sin2x = 0

⇔ \(3\sqrt{2}sin2x-cos2x=-1\)

Còn lại tự giải

7, \(cos\left(2x+\dfrac{\pi}{4}\right)+cos\left(2x-\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}\left(1-sinx\right)\)

⇔ \(2cos2x.cos\dfrac{\pi}{4}+4sinx=2+\sqrt{2}\left(1-sinx\right)\)

⇔ \(\sqrt{2}cos2x+4sinx=2+\sqrt{2}-\sqrt{2}sinx\)

Dùng công thức : cos2x = 1 - 2sin²x đưa về phương trình bậc 2 ẩn sinx

\(cosx.cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)=\frac{1}{2}cosx\left(cos\frac{2\pi}{3}+cos2x\right)=-\frac{1}{4}cosx+\frac{1}{2}cosx.cos2x\)

\(=-\frac{1}{4}cosx+\frac{1}{4}\left(cos3x+cosx\right)=\frac{1}{4}cos3x\)

\(sin5x-2sinx\left(cos4x+cos2x\right)=sinx.cos4x+cosx.sin4x-2sinx.cos4x-2sinx.cos2x\)

\(=sin4x.cosx-cos4x.sinx-2sinx.cos2x=sin3x-2sinx.cos2x\)

\(=sinx.cos2x+cosx.sin2x-2sinx.cos2x\)

\(=sin2x.cosx-cos2x.sinx=sinx\)

1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)

\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)

\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)

Vậy...

2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)

\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)

\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)

\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)

Vậy...

3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)

\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)

\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)

Vậy...

4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)

\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)

\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)

Vậy...

5, Xem lại đề

6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)

\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)

Vậy...