\(A=\dfrac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\dfrac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\dfrac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=tan2x\)

a.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)

\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)

b.

\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)

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

c.

\(\Leftrightarrow\dfrac{1}{2}cos6x+\dfrac{1}{2}cos4x=\dfrac{1}{2}cos6x+\dfrac{1}{2}cos2x+4-3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(2cos^22x-1\right)=\dfrac{1}{2}cos2x+\dfrac{5}{2}+\dfrac{3}{2}cos2x\)

\(\Leftrightarrow cos^22x-2cos2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)

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

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

a) cosx - √3sinx = √2 ⇔ cosx - tansinx = √2

⇔ coscosx - sinsinx = √2cos ⇔ cos(x + ) =

⇔

b) 3sin3x - 4cos3x = 5 ⇔ sin3x - cos3x = 1.

Đặt α = arccos thì phương trình trở thành

cosαsin3x - sinαcos3x = 1 ⇔ sin(3x - α) = 1 ⇔ 3x - α = + k2π

⇔ x = , k ∈ Z (trong đó α = arccos).

c) Ta có sinx + cosx = √2cos(x - ) nên phương trình tương đương với

2√2cos(x - ) - √2 = 0 ⇔ cos(x - ) =

⇔

d) 5cos2x + 12sin2x -13 = 0 ⇔

Đặt α = arccos thì phương trình trở thành

cosαcos2x + sinαsin2x = 1 ⇔ cos(2x - α) = 1

⇔ x = + kπ, k ∈ Z (trong đó α = arccos).

a: ĐKXĐ: sin 2x<>1

=>2x<>pi/2+k2pi

=>x<>pi/4+kpi

\(\dfrac{cos2x}{sin2x-1}=0\)

=>cos2x=0

=>2x=pi/2+kpi

=>x=pi/4+kpi/2

Kết hợp ĐKXĐ, ta được:

x=3/4pi+k2pi hoặc x=7/4pi+k2pi

b: cos(sinx)=1

=>sin x=kpi

=>sin x=0

=>x=kpi

c: \(2\cdot sin^2x-1+cos3x=0\)

=>cos3x+cos2x=0

=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)

=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)

=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi

=>x=-pi+k2pi hoặc x=pi/5+k2pi/5

e: cos3x=-cos7x

=>cos3x=cos(pi-7x)

=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi

=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2