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!!
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
By Raelynn
Đáp án:
Giải thích các bước giải:
mk gửi ảnh r đó
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)`
`<=>` \(\left[ \begin{array}{l}3x+\dfrac{π}{2}=x+\dfrac{π}{3}+k2π\\3x+\dfrac{π}{2}=π-x-\dfrac{π}{3}+k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{-π}{12}+kπ\\x= \dfrac{π}{24}+k \dfrac{π}{2}\end{array} \right.\)