Đáp án:

 Câu 29: $C$

Câu 30: $A$

Câu 31: $A$

Câu 32: $A$

Câu 33: $C$

Câu 34: $C$

Câu 35: $A$

Câu 36:$A$

Câu 37: $A$

Giải thích các bước giải:

 Câu 29:

Ta có: $tan(3x+60^o)=m^2 \Rightarrow \dfrac{tan3x+tan60^o}{1-tan3x.tan30^o}=m^2$

$\Rightarrow tan3x+tan60^o=m^2(1-tan3x.tan30^o)$

$\Rightarrow tan3x+\sqrt{3}=m^2(1-\sqrt{3}.tan3x)$

$\Rightarrow (1+\sqrt{3}.m^2).tan3x=m^2-\sqrt{3}$

Phương trình có nghiệm khi $1+\sqrt{3}.m^2 \neq 0 \Rightarrow m^2 \neq -\dfrac{1}{\sqrt{3}}$ ( luôn đúng)

$\Rightarrow m \in R$

Câu 30:

Ta có $ cos2x=-cos(x+\dfrac{\pi}{2})$

$\Rightarrow cos2x=cos(\pi-x-\dfrac{\pi}{2})$

$\Rightarrow cos2x=cos(\dfrac{\pi}{2}-x)$

$\left[ \begin{array}{l}2x=\dfrac{\pi}{2}-x+k2\pi\\2x=-\dfrac{\pi}{2}+x+k2\pi\end{array} \right.$

$\left[ \begin{array}{l}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{2}+k2\pi\end{array} \right.$

Do $x \in (0;10\pi)$ nên $\left[ \begin{array}{l}0<x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}<10\pi\\0<x=-\dfrac{\pi}{2}+k2\pi<10\pi\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}\dfrac{-\pi}{6}<\dfrac{k2\pi}{3}<\dfrac{59\pi}{6}\\\dfrac{\pi}{2}<k2\pi<\dfrac{21\pi}{2}\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}\dfrac{-1}{4}<k<\dfrac{59}{4}\\\dfrac{1}{4}<k<\dfrac{21}{4}\end{array} \right.$

$\Rightarrow k \in \{0;1;2;3;.....;14\}$

Vậy có $14$ giá trị

Câu 31:

Ta có: $2sinx-\sqrt{3}=0$

$\Rightarrow sinx=\dfrac{\sqrt{3}}{2}$

$\Rightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}0 \leq x=\dfrac{\pi}{3}+k2\pi \leq 2\pi\\0 \leq x=\dfrac{2\pi}{3}+k2\pi\leq 2\pi\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}-\dfrac{\pi}{3}\leq k2\pi \leq \dfrac{5\pi}{3}\\-\dfrac{2\pi}{3}\leq k2\pi \leq \dfrac{4\pi}{3}\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}-\dfrac{1}{6}\leq k \leq \dfrac{5}{6}\\-\dfrac{1}{3}\leq k \leq \dfrac{2}{3}\end{array} \right.$

$\Rightarrow k=0$

Vậy có $1$ giá trị

Câu 32:

Ta có: $mtanx-\sqrt{3}=0 \Rightarrow mtanx=\sqrt{3}$

Để phương trình có nghiệm thì $m \neq 0$

Câu 33:

Ta có: $2cos^2 x +m -1=0 \Rightarrow (2cos^2 x-1)+m=0$

$\Rightarrow cos2x+m=0 \Rightarrow cos2x=-m$

Ta có $-1\leq cos2x \leq 1 \Rightarrow -1 \leq -m \leq 1$

$\Rightarrow -1 \leq m \leq 1 \Rightarrow m\in \{-1;0;1\}$

Câu 34:

Ta có: $sin^6x+ cos^6x+3sinx.cosx-m+2=0$

$\Rightarrow (sin^2x+cos^2x)((sin^2x+cos^2x)^2-3sin^2x.cos^2x)+3sinx.cosx-m+2=0$

$\Rightarrow 1.(1^2-\dfrac{3}{4}sin^22x)+\dfrac{3}{2}sin^22x-m+2=0$

$\Rightarrow 1-\dfrac{3}{4}.sin^22x+\dfrac{3}{2}sin2x-m+2=0$

$\Rightarrow 4m=-3sin^22x+6sin2x+12$

Đặt $t=sin 2x, t \in$ [$-1;1$]

Xét $f(t) =-3t^2+6t+12$

Ta có: $f(-1)=3; f(1)=15$

$\Rightarrow 3 \leq 4m \leq 15 \Rightarrow \dfrac{3}{4} \leq m \leq \dfrac{15}{4}$

Vậy $ab=\dfrac{3}{4}.\dfrac{15}{4}=\dfrac{75}{16}$

Câu 35:

Phương trình có nghiệm khi $1^2+m^2 \geq (\sqrt{10})^2 \Rightarrow m^2 \geq 9$

$\Rightarrow \left[ \begin{array}{l}m\geq 3\\m \leq -3\end{array} \right.$

Câu 36:

Ta có: $cos2x+sin x=\sqrt{3}(cosx-sin2x) \Rightarrow cos2x+\sqrt{3}sin2x=\sqrt{3}cosx-sinx$

$\Rightarrow \dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x=\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx$

$\Rightarrow sin(\dfrac{\pi}{6}+2x)=sin(\dfrac{\pi}{3}-x)$

$\Rightarrow \left[ \begin{array}{l}\dfrac{\pi}{6}+2x=\dfrac{\pi}{3}-x+k2\pi\\\dfrac{\pi}{6}+2x=\pi-\dfrac{\pi}{3}+x-k2\pi\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{-3\pi}{2}-k2\pi\end{array} \right.$

Câu 37:

$(2cosx-1)(2cos2x+2cosx-m)=3-4sin^2x$

$\Rightarrow (2cosx-1)(2cos2x+2cosx-m)=(2cosx-1)(2cosx+1)$

$\Rightarrow (2cosx-1)(4cos^2x-2+2cosx-m)-(2cosx-1)(2cosx+1)=0$

$\Rightarrow (2cosx-1)(4cos^2x-2+2cosx-m-2cosx-1)=0$

$\Rightarrow (2cosx-1)(4cos^2x-3-m)=0$

$\Rightarrow \left[ \begin{array}{l}2cosx-1=0\\4cos^2x-3-m=0\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}cosx=\dfrac{1}{2}(1)\\cos^2x=\dfrac{m+3}{4}(2)\end{array} \right.$

$(1) \Leftrightarrow cosx=\dfrac{1}{2}$ có hai nghiệm thuộc [$\dfrac{-\pi}{2};\dfrac{\pi}{2}$]

$\Rightarrow$ Phương trình có nghiệm thuộc [$\dfrac{-\pi}{2};\dfrac{\pi}{2}$] $\Leftrightarrow (2)$ vô nghiệm hoặc $(2) \Leftrightarrow cosx=\dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l}\dfrac{m+3}{4}>1\\\dfrac{m+3}{4}<0\\\dfrac{m+3}{4}=\dfrac{1}{4}\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}m>1\\m<-3\\m=-2\end{array} \right.$

Vậy có $7$ giá trị m thỏa mãn

Câu 38: Không rõ đề