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Question Number 138725 by mathsuji last updated on 17/Apr/21

$$Solveforrealnumbers$$$$\frac{sin\left(sinx\right)}{sinx}+\frac{cos\left(cosx\right)}{cosx}=1$$

Answered by mr W last updated on 18/Apr/21

$$t=\sin x$$$$\cos x=\pm \sqrt{1-t^2}$$$$-1⩽t⩽1$$$$f\left(t\right)=\frac{\sin t}{t}+\frac{\cos \sqrt{1-t^2}}{\sqrt{1-t^2}}⩾f\left(0\right)=1+\cos 1>1$$$$\Rightarrow nosolutionforf\left(t\right)=1$$$$g\left(t\right)=\frac{\sin t}{t}-\frac{\cos \sqrt{1-t^2}}{\sqrt{1-t^2}}⩽g\left(0\right)=1-\cos 1<1=$$$$\Rightarrow nosolutionforg\left(t\right)=1$$$$$$$$\Rightarrow nosolutionfor$$$$\frac{sin\left(sinx\right)}{sinx}+\frac{cos\left(cosx\right)}{cosx}=1!$$$$$$$$\frac{sin\left(sinx\right)}{sinx}+\frac{cos\left(cosx\right)}{cosx}=ahasroots$$$$onlyifa⩾1+\cos 1ora⩽1-\cos 1$$

Commented by mathsuji last updated on 20/Apr/21

$$THANKSSIR$$

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