1) $\cos (2x)-\cos (x)-3 \sin (x)-2=0$
`⇒ 2 sqrt(2) cos(x/2) sin(x/2 + π/4) [sqrt(2) sin(-x + π/4) - 2] = 0`
`⇒ 2 sqrt(2) cos(x/2) sin(x/2 + π/4) [sqrt(2) sin(-x + π/4) - 2] = 0`
`⇒ cos(x/2) sin(x/2 + π/4) [sqrt(2) sin(-x + π/4) - 2] = 0`
`⇒` \(\left[ \begin{array}{l}cos(\frac{x}{2}) = 0\\\sqrt2 sin(-x + \frac{\pi}{4}) - 2 = 0\\sin(\frac{x}{2} + \frac{π}{4}) = 0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}\frac{x}{2} = π n + \frac{π}{2}\\sin(-x + \frac{π}{4})=\sqrt2\\\frac{x}{2} + \frac{π}{4} = π n\end{array} \right.\) `(n \in Z)`
`⇒` \(\left[ \begin{array}{l}x = 2π n + \pi\\-x + \frac{π}{4} = π - sin^{-1}(\sqrt2) + 2 π n\\-x + \frac{π}{4} = 2 π n+ sin^{-1}(\sqrt2)\\\frac{x}{2} = π n- \frac{π}{4}\end{array} \right.\) `(n \in Z)`
`⇒` \(\left[ \begin{array}{l}x = 2π n + \pi\\x = -\frac{3 π}{4} + sin^{-1}(\sqrt2) - 2 π n\\x = \frac{\pi}{4} - sin^{-1}(\sqrt2) - 2 π n\\x = 2π n- \frac{π}{2}\end{array} \right.\) `(n \in Z)`
Vậy phương trình có tập nghiệm:
`S={2 π n + π ; -(3 π)/4 + sin^(-1)(sqrt(2)) - 2 π n ; x = π/4 - sin^(-1)(sqrt(2)) - 2 π n ; 2 π n - π/2 | n \in \mathbb{Z}}`
3) `\cos (2x)+3 \cos (x)+2=\sin (x)`
`⇒ 2 + 3 cos(x) + cos(2 x) - sin(x) = 0`
`⇒ 2 sqrt(2) cos(x/2) sin(-x/2 + π/4) [sqrt(2) sin(x + π/4) + 2] = 0 `
`⇒ cos(x/2) sin(-x/2 + π/4) [sqrt(2) sin(x + π/4) + 2] = 0 `
`⇒` \(\left[ \begin{array}{l}cos(\frac{x}{2}) = 0 \\sin(\frac{-x}{2}+ \frac{π}{4}) \\ \sqrt2 sin(x + \frac{\pi}{4}) + 2 = 0 \end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}\frac{x}{2} = π n + \frac{π}{2}\\\frac{-x}{2} + \frac{π}{4} = π n \\ \sqrt2 sin(x+ \frac{\pi}{4}) = -2 \end{array} \right.\) `(n \in Z)`
`⇒`\(\left[ \begin{array}{l}x = 2π n + \pi\\\frac{-x}{2} = π n - \frac{π}{4}\\ sin(x + \frac{\pi}{4}) = - \sqrt2 \end{array} \right.\) `(n \in Z)`
`⇒`\(\left[ \begin{array}{l}
x = 2π n + \pi\\x = \frac{π}{2}- 2π n \\x + \frac{π}{4} = π + sin^{-1}(\sqrt2) + 2 π n \\ x + \frac{π}{4} = 2 π n - sin^{-1}(\sqrt2) \end{array} \right.\) `(n \in Z)`
`⇒` \(\left[ \begin{array}{l}
x = 2π n + \pi
\\x = \frac{π}{2}- 2π n
\\x = \frac{3 π}{4} + sin^{-1}(\sqrt2) + 2 π n
\\ x = \frac{-π}{4} - sin^{-1}(\sqrt2) + 2 π n
\end{array} \right.\) `(n \in Z)`
Vậy phương trình có tập nghiệm:
`S={ 2 π n + π ; π/2 - 2 π n ; (3 π)/4 + sin^(-1)(sqrt(2)) + 2 π n ; -π/4 - sin^(-1)(sqrt(2)) + 2 π n | n \in \mathbb{Z}}`
