Question

[Trọng]

giải giúp e đi ạ

Answer 1

[Minhmetmoi]

Ở đây ta dùng công thức:

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

PT

\(\Leftrightarrow\sin\left(\dfrac{3x}{2}+\dfrac{\pi}{4}\right)=3\cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)\)

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

Đặt \(t=\dfrac{x}{2}\)(Mình đặt lại để dễ nhìn)

Pt trở thành:

\(\sin3t+\cos3t=3(\sin t-\cos t)\)

\(\Leftrightarrow\left(3\sin t-4\sin^3t\right)+\left(4\cos^3t-3\cos t\right)=3\left(\sin t-\cos t\right)\)

\(\Leftrightarrow\sin^3t-\cos^3t=0\)

\(\Leftrightarrow\left(\sin t-\cos t\right)\left(1+\dfrac{\sin2t}{2}\right)=0\)

\(\Leftrightarrow\cos\left(t+\dfrac{\pi}{4}\right)=0\) (Do \(1+\dfrac{\sin2t}{2}>0\))

\(\Leftrightarrow t=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

hay \(x=\dfrac{\pi}{2}+k2\pi\)

Answer 2

[Nguyễn Việt Lâm]

Đặt \(\dfrac{\pi}{4}-\dfrac{x}{2}=t\Rightarrow\dfrac{x}{2}=\dfrac{\pi}{4}-t\)

\(\Rightarrow\dfrac{\pi}{4}+\dfrac{3x}{2}=\dfrac{\pi}{4}+3\left(\dfrac{\pi}{4}-t\right)=\pi-3t\)

Phương trình trở thành:

\(sin\left(\pi-3t\right)=3sint\)

\(\Leftrightarrow sin3t=3sint\)

\(\Leftrightarrow3sint-4sin^3t=3sint\)

\(\Leftrightarrow sint=0\)

\(\Rightarrow t=k\pi\)

\(\Rightarrow\dfrac{\pi}{4}-\dfrac{x}{2}=k\pi\)

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