Question

a)\(\sin x+\cos x=\sqrt{2}sin\dfrac{\pi}{12}\)

b)\(\sin2x+\sqrt{3}\cos2x=2\sin\left(x+\dfrac{\pi}{6}\right)\)

Answer 1

a.

\(\Leftrightarrow\dfrac{1}{\sqrt{2}}sinx+\dfrac{1}{\sqrt{2}}cosx=sin\dfrac{\pi}{12}\)

\(\Leftrightarrow sinx.cos\dfrac{\pi}{4}+cosx.sin\dfrac{\pi}{4}=sin\dfrac{\pi}{12}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{12}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{12}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{11\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 2

`a)sin x +cos x=\sqrt{2}sin` `\pi/12`

`<=>1/\sqrt{2}sin x+1/\sqrt{2}cos x=sin` `\pi/12`

`<=>sin(x+\pi/4)=sin` `\pi/12`

`<=>[(x+\pi/4=\pi/12+k2\pi),(x+\pi/4=[11\pi]/12+k2\pi):}`

`<=>[(x=-\pi/6+k2\pi),(x=[2\pi]/3+k2\pi):}`   `(k in ZZ)`

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`b)sin 2x+\sqrt{3}cos 2x=2sin(x+\pi/6)`

`<=>1/2sin 2x+\sqrt{3}/2 cos 2x=sin(x+\pi/6)`

`<=>sin(2x+\pi/3)=sin(x+\pi/6)`

`<=>[(2x+\pi/3=x+\pi/6+k2\pi),(2x+\pi/3=[5\pi]/6-x+k2\pi):}`

`<=>[(x=-\pi/6+k2\pi),(x=\pi/6+k[2\pi]/3):}`   `(k in ZZ)`

Answer 3

a) \(\sin x+\cos x=\sqrt{2}\sin\dfrac{\pi}{12}\)

\(\Rightarrow\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}\sin\dfrac{\pi}{12}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{12}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{11\pi}{12}+k2\pi\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

Answer 4

b.

\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=sin\left(x+\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow sin2x.cos\dfrac{\pi}{3}+cos2x.sin\dfrac{\pi}{3}=sin\left(x+\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=x+\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Answer 5

b) \(\sin2x+\sqrt{3}\cos2x=2\sin\left(x+\dfrac{\pi}{6}\right)\)

\(\Rightarrow\dfrac{1}{2}\sin2x+\dfrac{\sqrt{3}}{2}\cos2x=\sin\left(x+\dfrac{\pi}{6}\right)\\ \Rightarrow\sin\left(2x+\dfrac{\pi}{3}\right)=\sin\left(x+\dfrac{\pi}{6}\right)\\ \Rightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=x+\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=-x+\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)