Giải phương trình thuần bậc nhất với sinx và cosx: 1)✓3sin5x+cos5x=-2sin11x 2)✓3cos5x-2sin3x.cos2x-sinx=0 3)cos2x-✓3sin2x-✓3cosx+sinx-4=0 4)sinx+✓3cos

Đáp án: 1, x = π/96 + kπ/8
hoặc x =-5 π/36 – kπ/3 (k € Z)

2, <=> [ x = π/18 – kπ/3 
——[ x = -π/6 – kπ/2 

3,<=> x = π/3 + kπ

 4, \(\left[ \begin{array}{l}x=\frac{-π}{12} + k2π\\x=\frac{5π}{12} + k2π\end{array} \right.\) 

Giải thích các bước giải:

1)√3.sin5x + 2sin11x + cos5x = 0
=> √3/2.sin5x + 1/2. cos5x = – sin11x
=> cos(π/6).sin5x + sin(π/6).cos5x = sin(-11x)
=> sin(5x + π/6) = sin(-11x)
=> 5x + π/6 = -11x + k2π
hoặc 5x + π/6 = π-(-11x) + k2π
=> x = π/96 + kπ/8
hoặc x =-5 π/36 – kπ/3 (k € Z)

2) <=> √3.cos5x – (sin5x + sinx) – sinx = 0 
<=> √3cos5x – sin5x – 2sinx = 0 
<=> √3cos5x – sin5x = 2sinx— (2)(dang asinf(x) + bcosx = c) 
thỏa đk a² + b² ≥ c² 

Chia 2 vế cua pt 2 cho √(a² + b²) = 2 
(2) <=> √3/2.cos5x – 1/2.sin5x = sinx 
<=> sin(π/3).cos5x – cos(π/3).sin5x = sinx 
<=> sin(π/3 – 5x) = sinx 

<=> [ π/3 – 5x = x + k2π 
——[ π/3 – 5x = π – x + k2π 

<=> [ x = π/18 – kπ/3 
——[ x = -π/6 – kπ/2 
(k thuộc Z) 
3)cos2x – √3sin2x – √3cosx + sinx – 4 = 0

<=> (1/2)cos2x – (√3/2)sin2x – (√3/2)cosx + (1/2)sinx – 2 = 0

<=> [cos2x.cos(π/3) – sin2x.sin(π/3)] – [cosx.cos(π/6) – sinx.sin(π/6)] – 2 = 0

<=> cos(2x + π/3) – cos(x + π/6) – 2 = 0

<=> 2cos²(x + π/6) – cos(x + π/6) – 3 = 0

<=> [cos(x + π/6) + 1].[2cos(x + π/6) – 3] = 0

<=> cos(x + π/6) + 1 = 0 ( Vì 2cos(x + π/6) – 3 < 0)

<=> x + π/6 = π/2 + kπ

<=> x = π/3 + kπ

4) sinx + √3cosx=√2

<=> $\frac{1}{2}$ sinx +$\frac{√3}{2}$cosx = $\frac{√2}{2}$ 
⇔ cos$\frac{π}{3}$sinx +sin$\frac{π}{3}$cosx = sin$\frac{π}{4}$ 
<=> sin( x+ $\frac{π}{3}$ ) = sin$\frac{π}{4}$ 
<=> \(\left[ \begin{array}{l}x + \frac{π}{3}=\frac{π}{4} + k2π\\x + \frac{π}{3}=π -\frac{π}{4} + k2π \end{array} \right.\) 
<=> \(\left[ \begin{array}{l}x=\frac{-π}{12} + k2π\\x=\frac{5π}{12} + k2π\end{array} \right.\)