<=> 2cos (3x/2)cos(x/2+pi/3)=0

<=>cos (3x/2)=0 hoặc cos (x/2+pi/3)=0

<=>3x/2=pi/2+kpi hoặc x/2+pi/3=pi/2+kpi (k thuộc z)

<=>x=pi/3+(2/3)kpi hoặc x=pi/3+2kpi (k thuộc z)

KL

Chứng minh các biểu thức đã cho không phụ thuộc vào x.

Từ đó suy ra f'(x)=0

a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;

b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;

c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0

d,f(x)=\(\frac{3}{2}\)=>f'(x)=0

\(\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)-1+3cos\left(x+\frac{\pi}{3}\right)+2=0\)

\(\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)+3cos\left(x+\frac{\pi}{3}\right)+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+\frac{\pi}{3}\right)=-1\\cos\left(x+\frac{\pi}{3}\right)=-\frac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

a) \(x=-45^0+k90^0,k\in\mathbb{Z}\)

b) \(x=-\dfrac{\pi}{6}+k\pi,k\in\mathbb{Z}\)

c) \(x=\dfrac{3\pi}{4}+k2\pi,k\in\mathbb{Z}\)

d) \(x=300^0+k540^0,k\in\mathbb{Z}\)

Bài 3. a) cos (x - 1) = ⇔ x - 1 = ±arccos + k2π

⇔ x = 1 ±arccos + k2π , (k ∈ Z).

b) cos 3x = cos 12⁰ ⇔ 3x = ±12⁰ + k360⁰ ⇔ x = ±4⁰ + k120⁰ , (k ∈ Z).

c) Vì = cos nên ⇔ cos() = cos ⇔ = ± + k2π ⇔

d) Sử dụng công thức hạ bậc (suy ra trực tiếp từ công thức nhan đôi) ta có

⇔ ⇔

⇔ ⇔

3.3 d)

\(\sin8x-\cos6x=\sqrt{3}\left(\sin6x+\cos8x\right)\\ \Leftrightarrow\sin8x-\sqrt{3}\cos8x=\sqrt{3}\sin6x+\cos6x\\ \Leftrightarrow\sin\left(8x-\dfrac{\pi}{3}\right)=\sin\left(6x+\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\pi-\left(6x+\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{7}\end{matrix}\right.\)

3.4 a)

\(2sin\left(x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(-x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \)

Chia hai vế cho \(\sqrt{2^2+4^2}=2\sqrt{5}\)

Ta được:

\(\dfrac{1}{\sqrt{5}}cos\left(x-\dfrac{\pi}{4}\right)+\dfrac{2}{\sqrt{5}}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3}{4}\\ \)

Gọi \(\alpha\) là góc có \(cos\alpha=\dfrac{1}{\sqrt{5}}\)và \(sin\alpha=\dfrac{2}{\sqrt{5}}\)

Phương trình tương đương:

\(cos\left(x-\dfrac{\pi}{4}-\alpha\right)=\dfrac{3}{4}\\ \Leftrightarrow x=\pm arscos\left(\dfrac{3}{4}\right)+\dfrac{\pi}{4}+\alpha+k2\pi\)