Giải pt: 4sinx.sin(x + $\frac{\pi }{3}$) . sin( x + $\frac{2\pi }{3}$ ) + cos3x = 1

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1. @Nap_

   4sin x.sin ( x + $\frac{\pi}{3}$  ) . sin ( $\frac{2\pi}{3}$ ) cos3x = 1

   ⇔ 2sin x ( cos ( -$\frac{\pi}{3}$  ) – cos ( 2x + $\pi$ ) ) + cos3x = 1

   ⇔ 2sin x ( $\frac{1}{2}$ + cos2x ) + cos3x = 1

   ⇔ sin x + sin3x + sin ( -x ) + cos x = 1

   ⇔ sin3x + cos3x = 1

   ⇔√2sin ( 3x + $\frac{\pi}{4}$ ) = 1

   ⇔ sin ( 3x + $\frac{\pi}{4}$ ) = sin$\frac{\pi}{4}$ 

   ⇔\(\left[ \begin{array}{l}x = k\frac{\pi}{3}\\x = \frac{n}{6}+k\frac{\pi}{3}\end{array} \right.\) 

   @LonelyTeam@

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2. Đáp án:

    

   Giải thích các bước giải:

   Đặt $ t = x + \frac{π}{3} ⇒ 3x = 3t – π ⇒ cos3x = cos(3t – π) = – cos3t$ 

   $ 4sinx.sin(x + \frac{2π}{3}) = 2[cos\frac{2π}{3} – cos(2x + \frac{2π}{3})] = – (1 + 2cos2t)$

   Thay vào $PT : – sint(1 + 2cos2t) – cos3t = 1$

   $ ⇔ – sint – 2sintcos2t – cos3t = 1$

   $ ⇔ – sint + (sint – sin3t) – cos3t = 1$

   $ ⇔ sin3t + cos3t = – 1 ⇔ sin(3t + \frac{π}{4}) = – \frac{\sqrt[]{2}}{2}$

   @ $ 3t + \frac{π}{4} = – \frac{π}{4} + k2π ⇔ t = – \frac{π}{6} + k\frac{2π}{3}$

   $ ⇔ x + \frac{π}{3} = – \frac{π}{6} + k\frac{2π}{3} ⇔ x = – \frac{π}{2} + k\frac{2π}{3}$

   @ $ 3t + \frac{π}{4} = – \frac{3π}{4} + k2π ⇔ t = – \frac{π}{3} + k\frac{2π}{3}$

   $ ⇔ x + \frac{π}{3} = – \frac{π}{3} + k\frac{2π}{3} ⇔ x = m\frac{2π}{3}$

    

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