c.

\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)

\(\Leftrightarrow2cos\left(x+12^0\right)=1\)

\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)

2.

Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:

\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)

\(\Rightarrow-1\le m\le\dfrac{1}{2}\)

a.

\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

b.

\(2x-10^0=arccot\left(4\right)+k180^0\)

\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)

2.

Phương trình \(sin\left(3x-27^o\right)=2m^2+m\) có nghiệm khi:

\(2m^2+m\in\left[-1;1\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m\le1\\2m^2+m\ge-1\end{matrix}\right.\)

\(\Leftrightarrow\left(m+1\right)\left(2m-1\right)\le0\)

\(\Leftrightarrow-1\le m\le\dfrac{1}{2}\)

b)

(sin2x + cos2x)cosx + 2cos2x - sinx = 0

⇔ cos2x (cosx + 2) + sinx (2cos2 x – 1) = 0

⇔ cos2x (cosx + 2) + sinx.cos2x = 0

⇔ cos2x (cosx + sinx + 2) = 0

⇔ cos2x  = 0

⇔ 2x =  + kπ ⇔ x =  + k  (k ∈ )

c) 

Đáp án:

x=+ k2

và x= +k2 (k∈Z)

Lời giải:

sin2x-cos2x+3sinx-cosx-1=0

⇔ 2sinxcosx-(1-2sin²x) +3sinx-cosx-1=0

⇔ 2sin²x+2sinxcosx+3sinx-cosx-2=0

⇔ (2sin²x+3sinx-2)+ cosx(2sinx-1)=0

⇔ (2sinx-1)(sinx+2)+cosx(2sinx-1)=0

⇔ (2sinx-1)(sinx+cosx+2)=0

⇔ sinx=

⇔ x=+ k2

hoặc x= +k2 (k∈Z)

(sinx+cosx+2)=0 (vô nghiệm do sinx+cosx+2=sin(x+)+2>0)

Nhầm, câu c

1. 

ĐKXĐ: \(x\ne k\pi\)

\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)

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

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

2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.

3.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

4.

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

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

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

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)

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

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

ĐK: \(x\ne k\pi\)

\(\dfrac{1+sin2x+cos2x}{1+cot^2x}=sinx.\left(sin2x+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{cos^2x+sin^2x}{sin^2x}}=sinx.\left(2sinx.cosx+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{1}{sin^2x}}=2sin^2x.\left(cosx+sinx\right)\)

\(\Leftrightarrow1+sin2x+cos2x=2cosx+2sinx\)

\(\Leftrightarrow1+2sinx.cosx+2cos^2x-1=2cosx+2sinx\)

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

\(\Leftrightarrow\left(cosx-1\right).sin\left(x+\dfrac{\pi}{4}\right)=0\)

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

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

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

a) f'(x) = - 3sinx + 4cosx + 5. Do đó

f'(x) = 0 <=> - 3sinx + 4cosx + 5 = 0 <=> 3sinx - 4cosx = 5

<=> sinx - cosx = 1. (1)

Đặt cos φ = , (φ ∈) => sin φ = , ta có:

(1) <=> sinx.cos φ - cosx.sin φ = 1 <=> sin(x - φ) = 1

<=> x - φ = + k2π <=> x = φ + + k2π, k ∈ Z.

b) f'(x) = - cos(π + x) - sin = cosx + sin.

f'(x) = 0 <=> cosx + sin = 0 <=> sin = - cosx <=> sin = sin

<=> = + k2π hoặc = π - x + + k2π

<=> x = π - k4π hoặc x = π + k, (k ∈ Z).

1: cos(2x+pi/6)=cos(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=3x-pi/3+k2pi

=>5x=pi/6+k2pi hoặc -x=-1/2pi+k2pi

=>x=pi/30+k2pi/5 hoặc x=pi-k2pi

2: sin(2x+pi/6)=sin(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=pi-pi/3+3x+k2pi

=>5x=pi/6+k2pi hoặc -x=2/3pi-pi/6+k2pi

=>x=pi/30+k2pi/5 hoặc x=-1/2pi-k2pi

1) \(cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{3}-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\3x-2x=\dfrac{\pi}{3}+\dfrac{\pi}{6}-k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\) \(\left(k\in N\right)\)

a: \(\Leftrightarrow sin\left(\dfrac{x}{3}-\dfrac{pi}{4}\right)=sinx\)

=>x/3-pi/4=x+k2pi hoặc x/3-pi/4=pi-x+k2pi

=>2/3x=-pi/4+k2pi hoặc 4/3x=5/4pi+k2pi

=>x=-3/8pi+k3pi hoặc x=15/16pi+k*3/2pi

b: =>(sin3x-sin2x)(sin3x+sin2x)=0

=>sin3x-sin2x=0 hoặc sin 3x+sin 2x=0

=>sin 3x=sin 2x hoặc sin 3x=sin(-2x)

=>3x=2x+k2pi hoặc 3x=pi-2x+k2pi hoặc 3x=-2x+k2pi hoặc 3x=pi+2x+k2pi

=>x=k2pi hoặc x=pi/5+k2pi/5 hoặc x=k2pi/5 hoặc x=pi+k2pi

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

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}cos4x+\dfrac{1}{2}sin2x=0\)

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

\(\Leftrightarrow...\)