a) Ta có: `4 cos^2(x) + 3sin(x) cos(x) - sin^2(x) = 3`
`<=> 1/2 [3 + 5 cos(2 x) + 3 sin(2 x)] = 3`
`<=> 3 + 5 cos(2 x) + 3 sin(2 x) = 6`
`<=> -3 + 5 cos(2 x) + 3 sin(2 x) = 0`
`<=> -3 + (5-5tan^2(x))/(tan^2(x) + 1) + (6 tan(x))/(tan^2(x) + 1) = 0`
Đặt `y = tan(x)`, ta được:
`-3 + (5-5 y^2)/(y^2 + 1) + (6 y)/(y^2 + 1) = 0`
`<=> {-3(y^2+1)}/(y^2 + 1) + 5/(y^2 + 1) - (5 y^2)/(y^2 + 1) + (6 y)/(y^2 + 1) = 0`
`<=> (-8y^2+6y+2)/(y^2+1)=0`
`<=> (4 y^2 - 3 y - 1)/(y^2 + 1) = 0`
`<=> 4 y^2 - 3 y - 1 = 0`
`<=>(y - 1) (4 y + 1) = 0`
`<=>` \(\left[ \begin{array}{l}y - 1 = 0 \\4 y + 1 = 0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}y=1\\y = \frac{-1}{4}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}tan(x)=1\\tan(x) = \frac{-1}{4}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = π n + \frac{π}{4} (n ∈ \mathbb{Z})\\x = π n - tan^{-1} (\frac{1}{4}) (n ∈ \mathbb{Z})\end{array} \right.\)
Vậy phương trình có tập nghiệm:
`S={\pi n+\frac{\pi}{4} ; \pi n-\tan ^{-1}(\frac{1}{4}) | n \in \mathbb{Z}}`
b) Ta có: `cosx - sqrt3.sinx - cos3x =0`
`<=> sin(x) [2 sin(2 x) - sqrt(3)] = 0`
`<=>` \(\left[ \begin{array}{l}sin(x)=0\\2 sin(2 x) - \sqrt3 = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = π n (n \in \mathbb{Z})\\2sin(2 x) = \sqrt3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = π n (n \in \mathbb{Z})\\sin(2 x) = \frac{\sqrt3}{2} \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = π n (n \in \mathbb{Z})\\2 x = 2 π n +\frac{ 2 π}{3} (n \in \mathbb{Z})\\ 2x = 2 π n +\frac{ π}{3} (n \in \mathbb{Z}) \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = π n (n \in \mathbb{Z})\\ x = π n +\frac{ π}{3} (n \in \mathbb{Z})\\ x = π n +\frac{ π}{6} (n \in \mathbb{Z}) \end{array} \right.\)
Vậy phương trình có tập nghiệm:
`S={π n ; π n +\frac{ π}{3} ; π n +\frac{ π}{6} | n \in \mathbb{Z}}`
