Giải theo hướng sincos cossin ý ạ B) 3sin3x-4cos3x=5 D) 5cos2x+12sin2x-13=0

Đáp án:

b) $\text{ x = $\dfrac{\alpha}{3}$ + $\dfrac{\pi}{6}$+k $\dfrac{2\pi}{3}$ (K∈Z)}$

d) $\text{ x = $\dfrac{\alpha}{2}$ + k$\pi$ (K∈Z)}$

Giải thích các bước giải:

$\text{b) 2sin3x- 4cos3x =5}$

⇔$\text{ $\dfrac{2}{5}$ sin3x - $\dfrac{4}{5}$cos3x=1 }$

⇔$\text{ sin3x.cos$\alpha$ - cos3x.sin$\alpha$=1 }$

⇔$\text{ sin(3x - $\alpha$)=1 }$

⇔$\text{ 3x - $\alpha$=$\dfrac{\pi}{2}+k2\pi$ }$

⇔$\text{ 3x = $\alpha$ + $\dfrac{\pi}{2}+k2\pi$ }$

⇒$\text{ x = $\dfrac{\alpha}{3}$ + $\dfrac{\pi}{6}$+k $\dfrac{2\pi}{3}$ (K∈Z)}$

$\text{d) 5cos2x + 12sin2x -13=0}$

⇔$\text{ 5cos2x + 12sin2x =13}$

⇔$\text{$\dfrac{5}{13}$ cos2x +$\dfrac{12}{13}$sin2x =1}$

⇔$\text{ cos2x.cos$\alpha$ +sin2x.sin$\alpha$ =1}$

⇔$\text{ cos(2x - $\alpha$)=1}$

⇔$\text{ 2x - $\alpha$ = k2$\pi$}$

⇒$\text{ x = $\dfrac{\alpha}{2}$ + k$\pi$ (K∈Z)}$