Question

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giải phương trình sau

a) \(\cos2x+3\sin x-2=0\)

b) \(\cos3x.\cos x-\cos4x=2-4\sin^2\left(\frac{\pi}{4}-\frac{3x}{2}\right)\)

Answer 1

[Nguyễn Nam Dương]

a) 

\(\Leftrightarrow1-2sin^2x+3sinx-2=0.\)

\(\Leftrightarrow-2sin^2x+3sinx-1=0.\)

\(\orbr{\begin{cases}sinx=1\\sinx=\frac{1}{2}\end{cases}}\)

\(\hept{\begin{cases}x=\frac{\text{π}}{2}+k2\text{π}\\x=\frac{\text{π}}{6}+k2\text{π}\\x=\frac{5\text{π}}{6}+k2\text{π}\end{cases}}\)

b) Đề sai , xem hộ lại đè

b) 

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

Answer 2

[Nguyễn Nam Dương]

Bổ sung câu b

\(cos3x.cosx-cos4x=2-4sin^2\left(\dfrac{\pi}{4}-\dfrac{3x}{2}\right)\)

\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x-cos4x=2cos\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x=2sin3x\)

\(\Leftrightarrow sin3x.sinx=2sin3x\)

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

\(\Leftrightarrow sin3x=0\)

\(\Leftrightarrow3x=k\pi\)

\(\Leftrightarrow x=\dfrac{k\pi}{3}\)