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Question Number 164511 by HongKing last updated on 18/Jan/22

sin∙(π sin x) - cos∙(π sin x) = 1 find x=?

$$\mathrm{sin}\centerdot\left(\pi\:\mathrm{sin}\:\mathrm{x}\right)\:-\:\mathrm{cos}\centerdot\left(\pi\:\mathrm{sin}\:\mathrm{x}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$

Answered by mindispower last updated on 18/Jan/22

sin(a)−cos(a)=(√2)sin(a−(π/4)) ⇔ withea=πsin(x) sin(a−(π/4))=(1/( (√2))) ⇔a−(π/4)=(π/4)+2kπ,((3π)/4)+2kπ sin(x)=1+2k∈{1,−1} x=(π/2)+kπ πsin(x)=(π/2)+2kπ sin(x)=(1/2)+2k, ⇒k=0 sin(x)=(1/2),x∈{(π/6)+2kπ,((5π)/6)+2kπ}

$${sin}\left({a}\right)−{cos}\left({a}\right)=\sqrt{\mathrm{2}}{sin}\left({a}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Leftrightarrow\:{withea}=\pi{sin}\left({x}\right) \\ $$$${sin}\left({a}−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Leftrightarrow{a}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi,\frac{\mathrm{3}\pi}{\mathrm{4}}+\mathrm{2}{k}\pi \\ $$$${sin}\left({x}\right)=\mathrm{1}+\mathrm{2}{k}\in\left\{\mathrm{1},−\mathrm{1}\right\} \\ $$$${x}=\frac{\pi}{\mathrm{2}}+{k}\pi \\ $$$$\pi{sin}\left({x}\right)=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi \\ $$$${sin}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}{k},\:\:\Rightarrow{k}=\mathrm{0} \\ $$$${sin}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}},{x}\in\left\{\frac{\pi}{\mathrm{6}}+\mathrm{2}{k}\pi,\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2}{k}\pi\right\} \\ $$$$ \\ $$

Commented by HongKing last updated on 20/Jan/22

thank you dear Sir cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$

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