1.

Chắc đề là \(sin\left[\pi sin2x\right]=1?\)

\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)

Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)

\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)

Thế vào (1)

\(\Rightarrow sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)

Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)

Thế vào (2):

\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\) chắc bạn tự giải tiếp được

3.

\(\Leftrightarrow2sin\left(x+84^0\right).cos\left(60^0\right)=cos20^0\)

\(\Leftrightarrow sin\left(x+84^0\right)=sin70^0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+84^0=70^0+k360^0\\x+84^0=110^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-14^0+k360^0\\x=26^0+k360^0\end{matrix}\right.\)

b, \(VT=\dfrac{1-sin2x}{1+sin2x}\)

\(=\dfrac{sin^2x+cos^2x-2sinx.cosx}{sin^2x+cos^2x+2sinx.cosx}\)

\(=\dfrac{\left(sinx-cosx\right)^2}{\left(sinx+cosx\right)^2}\)

\(=\dfrac{\left(\dfrac{sinx-cosx}{cosx}\right)^2}{\left(\dfrac{sinx+cosx}{cosx}\right)^2}\)

\(=\dfrac{\left(\dfrac{sinx}{cosx}-1\right)^2}{\left(\dfrac{sinx}{cosx}+1\right)^2}\)

\(=\dfrac{\left(tanx-tan\dfrac{\pi}{4}\right)^2}{\left(1+tanx.tan\dfrac{\pi}{4}\right)^2}\)

\(=tan^2\left(x-\dfrac{\pi}{4}\right)=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\)

\(\Leftrightarrow sinx+sin2x=0\)

\(\Leftrightarrow sin2x=-sinx=sin\left(-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-x+k2\pi\\2x=\pi+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)

a: 3/2pi<x<2pi

=>sin x<0

=>\(sinx=-\sqrt{1-\left(\dfrac{1}{6}\right)^2}=-\dfrac{\sqrt{35}}{6}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{6}\cdot\dfrac{-\sqrt{35}}{6}=\dfrac{-\sqrt{35}}{18}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{1}{36}-1=\dfrac{1}{18}-1=\dfrac{-17}{18}\)

\(tan2x=\dfrac{-\sqrt{35}}{18}:\dfrac{-17}{18}=\dfrac{\sqrt{35}}{17}\)

\(cot2x=1:\dfrac{\sqrt{35}}{17}=\dfrac{17}{\sqrt{35}}\)

b: \(sin\left(\dfrac{pi}{3}-x\right)\)

\(=sin\left(\dfrac{pi}{3}\right)\cdot cosx-cos\left(\dfrac{pi}{3}\right)\cdot sinx\)

\(=\dfrac{1}{2}\cdot\dfrac{-\sqrt{35}}{6}-\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{-\sqrt{35}-1}{12}\)

c: \(cos\left(x-\dfrac{3}{4}pi\right)\)

\(=cosx\cdot cos\left(\dfrac{3}{4}pi\right)+sinx\cdot sin\left(\dfrac{3}{4}pi\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{-\sqrt{2}}{2}+\dfrac{-\sqrt{35}}{6}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{-\sqrt{2}-\sqrt{70}}{12}\)

d: tan(pi/6-x)

\(=\dfrac{tan\left(\dfrac{pi}{6}\right)-tanx}{1+tan\left(\dfrac{pi}{6}\right)\cdot tanx}\)

\(=\dfrac{\dfrac{\sqrt{3}}{3}-\sqrt{35}}{1+\dfrac{\sqrt{3}}{3}\cdot\left(-\sqrt{35}\right)}\)

a: pi/2<x<pi

=>cosx<0

=>\(cosx=-\sqrt{1-\left(\dfrac{1}{5}\right)^2}=-\dfrac{2\sqrt{6}}{5}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{5}\cdot\dfrac{-2\sqrt{6}}{5}=\dfrac{-4\sqrt{6}}{25}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{24}{25}-1=\dfrac{48}{25}-1=\dfrac{23}{25}\)

\(tan2x=-\dfrac{4\sqrt{6}}{25}:\dfrac{23}{25}=-\dfrac{4\sqrt{6}}{23}\)

\(cot2x=1:\dfrac{-4\sqrt{6}}{23}=\dfrac{-23}{4\sqrt{6}}\)

b: \(sin\left(x-\dfrac{pi}{6}\right)=sinx\cdot cos\left(\dfrac{pi}{6}\right)-cosx\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=sinx\cdot\dfrac{\sqrt{3}}{2}-cosx\cdot\dfrac{1}{2}\)

\(=\dfrac{1}{5}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{-2\sqrt{6}}{5}\cdot\dfrac{1}{2}=\dfrac{\sqrt{3}+2\sqrt{6}}{10}\)

c: \(cos\left(x-\dfrac{pi}{3}\right)=cosx\cdot cos\left(\dfrac{pi}{3}\right)+sinx\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=-\dfrac{2\sqrt{6}}{5}\cdot\dfrac{1}{2}+\dfrac{1}{5}\cdot\dfrac{1}{2}=\dfrac{-2\sqrt{6}+1}{10}\)

d: \(tan\left(x-\dfrac{pi}{4}\right)=\dfrac{tanx-tan\left(\dfrac{pi}{4}\right)}{1+tanx\cdot tan\left(\dfrac{pi}{4}\right)}\)

\(=\dfrac{tanx-1}{1+tanx}\)

\(=\dfrac{\dfrac{1}{-2\sqrt{6}}-1}{1+\dfrac{1}{-2\sqrt{6}}}=\dfrac{-25-4\sqrt{6}}{23}\)

1.

\(\Leftrightarrow2sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-1\end{matrix}\right.\)

\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)

Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)

3. Làm tương tự câu 2

4.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

a) Để tính sin2x, cos2x, tan2x và cot2x, chúng ta cần biết giá trị của cosx trước đã. Theo như bạn đã cho, cosx = -1/4. Vậy sinx sẽ bằng căn bậc hai của 1 - cos^2(x) = căn bậc hai của 1 - (-1/4)^2 = căn bậc hai của 1 - 1/16 = căn bậc hai của 15/16 = sqrt(15)/4. Sau đó, chúng ta có thể tính các giá trị khác như sau: sin2x = (2sinx*cosx) = 2 * (sqrt(15)/4) * (-1/4) = -sqrt(15)/8 cos2x = (2cos^2(x) - 1) = 2 * (-1/4)^2 - 1 = 2/16 - 1 = -14/16 = -7/8 tan2x = sin2x/cos2x = (-sqrt(15)/8) / (-7/8) = sqrt(15) / 7 cot2x = 1/tan2x = 7/sqrt(15) b) Để tính sin(x + 5π/6), chúng ta có thể sử dụng công thức sin(a + b) = sin(a)cos(b) + cos(a)sin(b). Với a = x và b = 5π/6, ta có: sin(x + 5π/6) = sin(x)cos(5π/6) + cos(x)sin(5π/6) = sin(x)(-sqrt(3)/2) + cos(x)(1/2) = (-sqrt(3)/2)sin(x) + (1/2)cos(x) c) Để tính cos(π/6 - x), chúng ta sử dụng công thức cos(a - b) = cos(a)cos(b) + sin(a)sin(b). Với a = π/6 và b = x, ta có: cos(π/6 - x) = cos(π/6)cos(x) + sin(π/6)sin(x) = (√3/2)cos(x) + 1/2sin(x) d) Để tính tan(x + π/3), chúng ta có thể sử dụng công thức tan(a + b) = (tan(a) + tan(b))/(1 - tan(a)tan(b)). Với a = x và b = π/3, ta có: tan(x + π/3) = (tan(x) + tan(π/3))/(1 - tan(x)tan(π/3))

a: pi/2<x<pi

=>sin x>0

=>\(sinx=\sqrt{1-\left(-\dfrac{1}{4}\right)^2}=\dfrac{\sqrt{15}}{4}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{\sqrt{15}}{4}\cdot\dfrac{-1}{4}=\dfrac{-\sqrt{15}}{8}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{1}{16}-1=-\dfrac{7}{8}\)

\(tan2x=-\dfrac{\sqrt{15}}{8}:\dfrac{-7}{8}=\dfrac{\sqrt{15}}{7}\)

\(cot2x=1:\dfrac{\sqrt{15}}{7}=\dfrac{7}{\sqrt{15}}\)

b: sin(x+5/6pi)

=sinx*cos(5/6pi)+cosx*sin(5/6pi)

\(=\dfrac{\sqrt{15}}{4}\cdot\dfrac{-\sqrt{3}}{2}+\dfrac{1}{2}\cdot\dfrac{-1}{4}=\dfrac{-\sqrt{45}-1}{8}\)

c: cos(pi/6-x)

=cos(pi/6)*cosx+sin(pi/6)*sinx

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{-1}{4}+\dfrac{1}{2}\cdot\dfrac{\sqrt{15}}{4}=\dfrac{-\sqrt{3}+\sqrt{15}}{8}\)

d: tan(x+pi/3)

\(=\dfrac{tanx+tan\left(\dfrac{pi}{3}\right)}{1-tanx\cdot tan\left(\dfrac{pi}{3}\right)}\)

\(=\dfrac{-\sqrt{15}+\sqrt{3}}{1+\sqrt{15}\cdot\sqrt{3}}=\dfrac{-\sqrt{15}+\sqrt{3}}{1+3\sqrt{5}}\)