Hướng dẫn giải (chi tiết)

Điều kiện $\left\{\begin{array}{l}{x \neq \dfrac{\pi}{2}+k \pi} \\ {y \neq \dfrac{\pi}{2}+k \pi}\end{array}\right.$

Ta có 

$\left\{\begin{array}{l}{x+y=\dfrac{2 \pi}{3}} \\ {\tan x \cdot \tan y=3}\end{array} \Leftrightarrow\left\{\begin{array}{l}{y=\dfrac{2 \pi}{3}-x} \\ {\tan x \cdot \tan \left(\dfrac{2 \pi}{3}-x\right)=3 (2)}\end{array}\right.\right.$

$(2)\Leftrightarrow \tan x \cdot \dfrac{\tan \dfrac{2 \pi}{3}-\tan x}{1+\tan \dfrac{2 \pi}{3} \cdot \tan x}=3 \Leftrightarrow \tan ^{2} x-2 \sqrt{3} \tan x+3=0 \Leftrightarrow \tan x=\sqrt{3} \Leftrightarrow x=\dfrac{\pi}{3}+k \pi$

Suy ra $y=\dfrac{\pi}{3}-k \pi$

Hướng dẫn giải (chi tiết)

$\sin ^{2} x+2(m+1) \sin x-3 m(m-2)=0(*)$

Đặt $\sin x=t(-1 \leq t \leq 1)$ khi đó phương trình có dạng $t^{2}+2(m+1) t-3 m(m-2)=0(1)$

Ta có $\Delta^{\prime}=(m+1)^{2}+3 m(m-2)=4 m^{2}-4 m+1=(2 m-1)^{2} \geq 0 \forall m \in \mathbb{Z}$

$\mathrm{TH} 1 : \Delta^{\prime}=0 \Leftrightarrow m=\dfrac{1}{2}$ phương trình (1) có nghiệm $t=-m-1=\dfrac{-1}{2}-1=-\dfrac{3}{2}(k t m)$

$\mathrm{TH} 2 : \Delta^{\prime}>0 \Leftrightarrow m \neq \dfrac{1}{2}$ phương trình (1) có 2 nghiệm $\left[\begin{array}{l}{t_{1}=-m-1+2 m-1=m-2} \\ {t_{2}=-m-1-2 m+1=-3 m}\end{array}\right.$

Để $(*)$ có nghiệm thì (1) có nghiệm $-1 \leq t \leq 1$

$\Leftrightarrow\left[\begin{array}{l}{-1 \leq m-2 \leq 1} \\ {-1 \leq-3 m \leq 1}\end{array} \Leftrightarrow\left[\begin{array}{l}{1 \leq m \leq 3} \\ {-\dfrac{1}{3} \leq m \leq \dfrac{1}{3}}\end{array}\right.\right.$

Hướng dẫn giải (chi tiết)

$\begin{aligned} & 3 \sin 2 x+4 \cos 2 x+5 \cos 2017 x=0 \\ \Leftrightarrow & \dfrac{3}{5} \sin 2 x+\dfrac{4}{5} \cos 2 x+\cos 2017 x=0 \end{aligned}$

Đặt $\dfrac{3}{5}=\sin \alpha \Rightarrow \dfrac{4}{5}=\cos \alpha$ khi đó ta có 

$\begin{array}{l}{\sin 2 x \sin \alpha+\cos 2 x \cos \alpha=-\cos 2017 x} \\ {\Leftrightarrow \cos (2 x-\alpha)=\cos (\pi-2017 x)} \\ {\Leftrightarrow\left[\begin{array}{l}{2 x-\alpha=\pi-2017 x+k 2 \pi} \\ {2 x-\alpha=-\pi+2017 x+k 2 \pi}\end{array}\right.}\end{array}$

$\Leftrightarrow\left[\begin{array}{l}{2019 x=\pi+\alpha+k 2 \pi} \\ {-2015 x=-\pi+\alpha+k 2 \pi}\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{l}{x=\dfrac{\pi+\alpha}{2019}+\dfrac{k 2 \pi}{2019}} \\ {x=\dfrac{\pi-\alpha}{2015}-\dfrac{k 2 \pi}{2015}}\end{array} \quad(k \in \mathbb{Z})\right.$

Vậy phương trình có hai họ nghiệm.

Hướng dẫn giải (chi tiết)

Với $x \in\left(0 ; \dfrac{\pi}{2}\right)$ ta có $\left\{\begin{array}{l}{\sin x>0} \\ {\cos x>0}\end{array} \Rightarrow\left\{\begin{array}{l}{\tan x>0} \\ {\cot x>0}\end{array}\right.\right.$

Ta có $\tan x+\cot x=\tan x+\dfrac{1}{\tan x} \geq 2 \sqrt{\tan x \cdot \dfrac{1}{\tan x}}=2$

Phương trình có nghiệm $\Leftrightarrow m \geq 2$

Vậy $m \in\{2 ; 3 ; 4\}$ nên tổng các giá trị là 9

Hướng dẫn giải (chi tiết)

$(2 \sin x-1)(\sqrt{3} \tan x+2 \sin x)=3-4 \cos ^{2} x(*)$

Điều kiện $\cos x \neq 0 \Leftrightarrow x \neq \dfrac{\pi}{2}+k \pi$

$\begin{array}{l}{(*) \Leftrightarrow(2 \sin x-1) \cdot \dfrac{\sqrt{3} \sin x+2 \sin x \cos x}{\cos x}=3-4 \cos ^{2} x} \\ {\Leftrightarrow(2 \sin x-1)(\sqrt{3} \sin x+\sin 2 x)+\left(4 \cos ^{3} x-3 \cos x\right)=0} \\ {\Leftrightarrow 2 \sqrt{3} \sin ^{2} x-\sqrt{3} \sin x+2 \sin x \sin 2 x-\sin 2 x+\cos 3 x=0} \\ {\Leftrightarrow 2 \sqrt{3} \sin ^{2} x-\sqrt{3} \sin x+\cos x-\cos 3 x-\sin 2 x+\cos 3 x=0}\end{array}$

$\begin{array}{l}{\Leftrightarrow \sqrt{3} \sin x(2 \sin x-1)-\sin 2 x+\cos x=0} \\ {\Leftrightarrow \sqrt{3} \sin x(2 \sin x-1)-\cos x(2 \sin x-1)=0} \\ {\Leftrightarrow(2 \sin x-1)(\sqrt{3} \sin x-\cos x)=0}\end{array}$

$\Leftrightarrow\left[\begin{array}{l}{2 \sin x-1=0} (1) \\ {\sqrt{3} \sin x-\cos x=0 (2) }\end{array}\right.$

$(1) \Leftrightarrow \sin x=\dfrac{1}{2} \Leftrightarrow\left[\begin{array}{l}{x=\dfrac{\pi}{6}+k 2 \pi} \\ {x=\dfrac{5 \pi}{6}+k 2 \pi}\end{array}\right.$

$(2) \Leftrightarrow \sqrt{3} \sin x=\cos x \Leftrightarrow \sqrt{3} \tan x=1 \Leftrightarrow \tan x=\dfrac{1}{\sqrt{3}} \Leftrightarrow x=\dfrac{\pi}{6}+k \pi(T M)$

Hợp nghiệm (1) và (2) ta được $\left[\begin{array}{l}{x=\dfrac{\pi}{6}+k \pi} \\ {x=\dfrac{5 \pi}{6}+k 2 \pi}\end{array}(k \in \mathbb{Z})\right.$

Mà $x \in[0 ; 20 \pi] \Rightarrow x \in\left\{\dfrac{\pi}{6} ; \dfrac{\pi}{6}+\pi ; \ldots ; \dfrac{\pi}{6}+19 \pi ; \dfrac{5 \pi}{6} ; \dfrac{5 \pi}{6}+2 \pi ; \ldots \dfrac{5 \pi}{6}+18 \pi\right\}$

Vậy tổng các nghiệm là :

$\begin{aligned} & \dfrac{\pi}{6}+\dfrac{\pi}{6}+\pi+\dfrac{\pi}{6}+2 \pi+\ldots+\dfrac{\pi}{6}+19 \pi+\dfrac{5 \pi}{6}+\dfrac{5 \pi}{6}+2 \pi+\ldots+\dfrac{5 \pi}{6}+18 \pi \\=& 20 . \dfrac{\pi}{6}+(1+2+3+\ldots+19) \pi+\dfrac{5 \pi}{6} \cdot 10+2 \pi(1+2+\ldots+9)=\dfrac{875 \pi}{3} \end{aligned}$

Hướng dẫn giải (chi tiết)

Ta có $\sin ^{2015} x-\cos ^{2016} x=2\left(\sin ^{2017} x-\cos ^{2018} x\right)+\cos 2 x$

$\begin{array}{l}{\Leftrightarrow \sin ^{2015} x\left(1-2 \sin ^{2} x\right)+\cos ^{2016} x\left(2 \cos ^{2} x-1\right)=\cos 2 x} \\ {\Leftrightarrow \sin ^{2015} x \cdot \cos 2 x+\cos ^{2016} x \cdot \cos 2 x=\cos 2 x \Leftrightarrow\left[\begin{array}{l}{\cos 2 x=0} \\ {\sin ^{2015} x+\cos ^{2016} x=1}\end{array}\right.}\end{array}$

Với $\cos 2 x=0 \Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k \pi}{2}, k \in \mathbb{Z}$

Vì $x \in[-10 ; 30] \Rightarrow-10 \leq \dfrac{\pi}{4}+k \dfrac{\pi}{2} \leq 30 \Leftrightarrow-\dfrac{20}{\pi}-\dfrac{1}{2} \leq k \leq \dfrac{60}{\pi}-\dfrac{1}{2} \Rightarrow-6 \leq k \leq 18$

Với $\sin ^{2015} x+\cos ^{2016} x=1$ . Ta có $\sin ^{2015} x \leq \sin ^{2} x ; \cos ^{2016} x \leq \cos ^{2} x$

Do đó $1=\sin ^{2015} x+\cos ^{2016} x \leq \sin ^{2} x+\cos ^{2} x=1$nên $\left[\begin{array}{l}{\sin x=0, \cos x=\pm 1} \\ {\sin x=1, \cos x=0}\end{array}\right.$

Nếu $\sin x=0 \Leftrightarrow x=k \pi, k \in \mathbb{Z}$

Vì $x \in[-10 ; 30] \Rightarrow-10 \leq k \pi \leq 30 \Leftrightarrow \dfrac{-10}{\pi} \leq k \leq \dfrac{30}{\pi} \Rightarrow-3 \leq k \leq 9$

Nếu $\sin x=1 \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi, k \in \mathbb{Z}$

Vì $x \in[-10 ; 30] \Rightarrow-10 \leq \dfrac{\pi}{2}+k 2 \pi \leq 30 \Leftrightarrow-\dfrac{5}{\pi}-\dfrac{1}{4} \leq k \leq \dfrac{15}{\pi}-\dfrac{1}{4} \Rightarrow-1 \leq k \leq 4$

Vậy có 44 nghiệm thỏa mãn điều kiện

Hướng dẫn giải (chi tiết)

Ta có:  $-1 \leq \sin x \leq 1 ;-1 \leq \cos x \leq 1$

$\begin{array}{l}{\sin ^{3} x+\sin ^{2} x=\sin ^{2} x(\sin x+1) \geq 0} \\ {\sin ^{3} x-\sin ^{2} x=\sin ^{2} x(\sin x-1) \leq 0}\end{array}$

Do đó $-\sin ^{2} x \leq \sin ^{3} x \leq \sin ^{2} x$

Tương tự $-\cos ^{2} x \leq \cos ^{3} x \leq \cos ^{2} x$

$\Rightarrow-2 \sin ^{2} x-\cos ^{2} x \leq y \leq 2 \sin ^{2} x+\cos ^{2} x$

Mà $\left\{\begin{array}{l}{-2 \sin ^{2} x-\cos ^{2} x=-1-\sin ^{2} x \geq-1-1=-2} \\ {2 \sin ^{2} x+\cos ^{2} x=1+\sin ^{2} x \leq 1+1=2}\end{array}\right.$ nên $-2 \leq y \leq 2$

Vậy $M^{2}+m^{2}=8$

Hướng dẫn giải (chi tiết)

Ta có :

$\dfrac{{\sin \left( {x - y} \right)}}{{\sin x.\sin y}} = \dfrac{{\sin x.\cos y - \sin y.\cos x}}{{\sin x.\sin y}} = \cot y - \cot x$

Đặt $P=\dfrac{1}{\sin 45^{\circ} \cdot \sin 46^{\circ}}+\dfrac{1}{\sin 46^{\circ} \cdot \sin 47^{\circ}}+\cdots+\dfrac{1}{\sin 134^{\circ} \cdot \sin 135^{\circ}}$

$\begin{array}{l}{\Rightarrow \sin 1^{\circ} \cdot P=\dfrac{\sin 1^{\circ}}{\sin 45^{\circ} \cdot \sin 46^{\circ}}+\dfrac{\sin 1^{\circ}}{\sin 46^{\circ} \cdot \sin 47^{\circ}}+\cdots+\dfrac{\sin 1^{\circ}}{\sin 134^{\circ} \cdot \sin 135^{\circ}}} \\ {\Rightarrow \sin 1^{\circ} \cdot P=\cot 45^{\circ}-\cot 46^{\circ}+\cot 46^{\circ}-\cot 47^{\circ}+\ldots+\cot 134^{\circ}-\cot 135^{\circ}}\end{array}$

$\Rightarrow \sin 1^{\circ} . \mathrm{P}=\cot 45^{\circ}-\cot 135^{\circ}=2 \Rightarrow \mathrm{P}=\dfrac{2}{\sin 1^{\circ}} \Rightarrow \mathrm{n}=1$

Hướng dẫn giải (chi tiết)

Phương trình tương đương với $\sin 5 x=1-2 \cos ^{2} x \Leftrightarrow \sin 5 x=-\cos 2 x$

$\Leftrightarrow \sin 5 x=\sin \left(2 x-\dfrac{\pi}{2}\right) \Leftrightarrow\left[\begin{array}{l}{x=-\dfrac{\pi}{6}+k \dfrac{2 \pi}{3}} \\ {x=\dfrac{3 \pi}{14}+k \dfrac{2 \pi}{7}}\end{array}\right.$

Vậy nghiệm dương nhỏ nhất là $\dfrac{3 \pi}{14} \Rightarrow\left\{\begin{array}{l}{a=3} \\ {b=14}\end{array} \Rightarrow S=17\right.$