1) 4$tan^{2}3x$ – $\frac{1}{cos^{2}3x }$ = 2 2) $sin2x +cos2x$ =$\sqrt[]{2}sin3x$ 3) $cos3x – sinx$ = $\sqrt[]{3}(cosx – sin3x)$ P/s : Mọi ngườ

Bởi

1) 4$tan^{2}3x$ – $\frac{1}{cos^{2}3x }$ = 2
2) $sin2x +cos2x$ =$\sqrt[]{2}sin3x$
3) $cos3x – sinx$ = $\sqrt[]{3}(cosx – sin3x)$
P/s : Mọi người giúp mk vs ạ, mk cảm ơn!!

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  2. 1. `4tan^2 3x – 1/(cos^2 3x) = 2`

    `<=> 4tan^2 3x – 1 – tan^2 3x = 2`

    `<=>3tan^2 3x = 3`

    `<=> tan^2 3x = 1`

    `<=> 3x = π/4 + kπ`

    `<=> x = π/12 + k π/3`

    2. `sin2x + cos2x = \sqrt2 sin3x`

    `<=>\sqrt2/2 sin2x + \sqrt2/2 cos 2x= sin3x`

    `<=> sin (2x+π/4) = sin3x`

    `<=>` \(\left[ \begin{array}{l}2x+\dfrac{π}{4}=3x+k2π\\2x+\dfrac{π}{4}=π-3x+k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x=\dfrac{π}{4}+k2π\\x=\dfrac{3π}{20}+k \dfrac{2π}{5}\end{array} \right.\) 

    3. `cos3x – sinx = \sqrt3(cosx – sin3x)`

    `<=> cos3x-sinx = \sqrt3cosx – \sqrt3 sin3x`

    `<=> cos3x + \sqrt3 sin3x = \sqrt3 cosx + sinx`

    `<=> 1/2 cos3x + \sqrt3/2 sin3x = \sqrt3/2 cosx + 1/2 sinx`

    `<=> sin(3x+π/2) = sin(x+π/3)`

    `<=>` \(\left[ \begin{array}{l}3x+\dfrac{π}{2}=x+\dfrac{π}{3}+k2π\\3x+\dfrac{π}{2}=π-x-\dfrac{π}{3}+k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x=\dfrac{-π}{12}+kπ\\x= \dfrac{π}{24}+k \dfrac{π}{2}\end{array} \right.\) 

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