## 2cos$^{2}$ (x+$\frac{\pi }{3}$ ) + 5sin(x + $\frac{\pi }{3}$ ) -4=0

Question

2cos$^{2}$ (x+$\frac{\pi }{3}$ ) + 5sin(x + $\frac{\pi }{3}$ ) -4=0

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2 tuần 2021-07-12T10:34:41+00:00 2 Answers 1 views 0

1. Đáp án: x=\frac{+π}{6}+k2π;x=\frac{π}{2}+k2π

Giải thích các bước giải:

2cos^2 (x+π/3) + 5sin (x+π/3) – 4 =0

<=> 2(1-sin^2 (x+π/3)) + 5sin(x+π/3) – 4=0

<=>  2sin^2 (x+π/3) * 5sin(x+π/3) + 2 =0

Đặt t= sin (x+π/3) ( -1 ≤t≤t), được:

2t^2 -5t+2=0 <=> $$\left[ \begin{array}{l}t=2(L)\\t=\dfrac{1}{2}\end{array} \right.$$

Với t = 1/2 có : sin (x+π/3) = sin (π/6)

<=> $$\left[ \begin{array}{l}x+\dfrac{π}{3}=\dfrac{π}{6}+k2π\\x+\dfrac{π}{3}=π-\dfrac{π}{6}+k2π\end{array} \right.$$

<=> $$\left[ \begin{array}{l}x=\dfrac{-π}{6}+k2π\\x=\dfrac{π}{2}+k2π\end{array} \right.$$

Vậy PT có 2 họ nghiệm như trên.

2. 2cos² (x + π/3) + 5sin (x + π/3) – 4 = 0

Đặt t = x + π/3

=> 2cos² t + 5sin t – 4 = 0

<=> 2 – 2sin² t + 5sin t – 4 = 0

<=> $$\left[ \begin{array}{l}sin t = 2 (l)\\sin t = \dfrac{1}{2}\end{array} \right.$$

=> sin t = 1/2

<=> sin (x + π/3) = sin (π/6)

<=> $$\left[ \begin{array}{l}x + \dfrac{π}{3} = \dfrac{π}{6} + k2π\\x + \dfrac{π}{3} = \dfrac{5π}{6} + k2π\end{array} \right.$$

<=> $$\left[ \begin{array}{l}x = -\dfrac{π}{6} + k2π\\x = \dfrac{π}{2} + k2π\end{array} \right.$$ (k ∈ ZZ)