a/ ĐKXĐ: \(cos2x\ne0\)

\(\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)

Pt tương đương:

\(\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\2cosx+\sqrt{2}=0\\sin2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\cosx=cos\left(\frac{3\pi}{4}\right)\\2x=k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=\frac{3\pi}{4}+k2\pi\left(l\right)\\x=-\frac{3\pi}{4}+k2\pi\left(l\right)\\x=\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x=\frac{k\pi}{2}\)

b/

ĐKXĐ: \(x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)

\(\Leftrightarrow tan2x.sinx+3sinx-\sqrt{3}tan2x-3\sqrt{3}=0\)

\(\Leftrightarrow sinx\left(tan2x+3\right)-\sqrt{3}\left(tan2x+3\right)=0\)

\(\Leftrightarrow\left(sinx-\sqrt{3}\right)\left(tan2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\sqrt{3}>1\left(vn\right)\\tan2x=-3\end{matrix}\right.\)

\(\Rightarrow2x=arctan\left(-3\right)+k\pi\)

\(\Rightarrow x=\frac{arctan\left(-2\right)}{2}+\frac{k\pi}{2}\)

c/

ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x+\frac{3\pi}{4}\right)\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\frac{3\pi}{4}\ne k\pi\\2x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ne-\frac{3\pi}{4}+k\pi\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)

Pt tương đương:

\(cos^22x=sin^2\left(x+\frac{3\pi}{4}\right)\)

\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{3\pi}{2}\right)\)

\(\Leftrightarrow cos4x=-cos\left(2x+\frac{3\pi}{2}\right)=cos\left(2x+\frac{\pi}{2}\right)\)

\(\Rightarrow\left[{}\begin{matrix}4x=2x+\frac{\pi}{2}+k2\pi\\4x=-2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)

\(\dfrac{cos4x}{cos2x}=tan2x\). ĐKXĐ : \(x\ne\dfrac{\pi}{4}+k.\dfrac{\pi}{2}\), k là số nguyên (tức là sin2x khác 1 và -1)

⇒ cos4x = sin2x

⇔ 1 - 2sin²2x = sin2x

⇔ 2sin²2x + sin2x - 1 = 0 

⇔ \(\left[{}\begin{matrix}sin2x=-1\left(/\right)\\sin2x=\dfrac{1}{2}\left(V\right)\end{matrix}\right.\)

Mà x ∈ \(\left(0;\dfrac{\pi}{2}\right)\)

⇒ \(\left[{}\begin{matrix}x=\dfrac{\pi}{6}\\x=\dfrac{\pi}{3}\end{matrix}\right.\)

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

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

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

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)

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

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

a: pi<x<3/2pi

=>cosx<0

=>\(cosx=-\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)

\(tanx=\dfrac{-3}{5}:\dfrac{-4}{5}=\dfrac{3}{4}\)

cot x=1:3/4=4/3

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{-3}{5}\cdot\dfrac{-4}{5}=\dfrac{24}{25}\)

\(cos2x=1-2\cdot sin^2x=1-2\cdot\left(-\dfrac{3}{5}\right)^2=\dfrac{7}{25}\)

\(tan2x=\dfrac{24}{25}:\dfrac{7}{25}=\dfrac{24}{7}\)

cot 2x=1:24/7=7/24

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

\(=\dfrac{-3}{5}\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{-4}{5}=\dfrac{-3-4\sqrt{3}}{10}\)

1.

\(\Leftrightarrow2cos2x+sinx-sin3x=0\)

\(\Leftrightarrow2cos2x-2cos2x.sinx=0\)

\(\Leftrightarrow2cos2x\left(1-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=1\end{matrix}\right.\)

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

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

2.

\(cos^2x+\left(sin3x-1\right)\left(1-cos\left(\dfrac{\pi}{2}-x\right)\right)=0\)

\(\Leftrightarrow1-sin^2x+\left(sin3x-1\right)\left(1-sinx\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx\right)+\left(sin3x-1\right)\left(1-sinx\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx+sin3x-1\right)=0\)

\(\Leftrightarrow2\left(1-sinx\right)sin2x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sin2x=0\\cosx=0\end{matrix}\right.\)

\(\Leftrightarrow sin2x=0\)

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

a/ \(\frac{\pi}{6}< x< \frac{\pi}{3}\Rightarrow cosx>0\)

\(cos^2x=\frac{1}{1+tan^2x}=\frac{1}{10}\)

\(cotx=\frac{1}{tanx}=\frac{1}{3}\)

Thay số và bấm máy

b/ \(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\tana< 0\end{matrix}\right.\)

\(sina=\sqrt{1-cos^2a}=\frac{3}{5}\)

\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)

\(A=\frac{6sina.cosa-\frac{2tana}{1-tan^2a}}{cosa-\left(2cos^2a-1\right)}\)

Thay số và bấm máy

c/ \(\frac{3\pi}{2}< x< 2\pi\Rightarrow\left\{{}\begin{matrix}cosx>0\\sinx< 0\end{matrix}\right.\)

\(cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{1}{\sqrt{5}}\)

\(sinx=cosx.tanx=-\frac{2}{\sqrt{5}}\)

\(B=\frac{cos^2x+2sinx.cosx}{\frac{2tanx}{1-tan^2x}-\left(2cos^2x-1\right)}\)

Thay số

1: 90<x<180

=>cosx<0

=>\(cosx=-\sqrt{1-\left(\dfrac{4}{5}\right)^2}=-\dfrac{3}{5}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{4}{5}\cdot\dfrac{-3}{5}=\dfrac{-24}{25}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{9}{25}-1=-\dfrac{7}{25}\)

\(tan2x=\dfrac{-24}{25}:\dfrac{-7}{25}=\dfrac{24}{7}\)

2: 0<x<90

=>cosx>0

=>\(cosx=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{3}{4}-1=\dfrac{6}{4}-1=\dfrac{2}{4}=\dfrac{1}{2}\)