Solving esin⁡x+ecos⁡x=e+1

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delai59qk · Accepted Answer

I&#039;ll place here a proof of the statement: the only real roots of the equationesin⁡x+ecos⁡x=e+1are x=2πk, x=π2+2πk, k∈ZFrom 2π- periodicity of sin⁡ and cos⁡ it is sufficient to show that the only roots from [0,2π) are x=0 and x=π2. For π2&amp;lt;x&amp;lt;2π one of functions sin or cos is negative, hence in the sum esin⁡x+ecos⁡x one summand is &amp;lt;1. The other is always ≤e and the value e+1 can&#039;t be obtained.It remains to consider x∈[0,π2]. Let f(x)=esin⁡x+ecos⁡x, 0≤x≤π2. Then for 0&amp;lt;x&amp;lt;π2f′(x)=cos⁡xesin⁡x−sin⁡xecos⁡x=sin⁡xcos⁡x(esin⁡xsin⁡x−ecos⁡xcos⁡x)=sin⁡xcos⁡x(g(sin⁡x)−g(cos⁡x))where g(t)=ett. For 0&amp;lt;t&amp;lt;1 the function g is decreasing, as g′(t)=et(t−1)t2. Thenf′(x)&amp;gt;0⇔g(sin⁡x)&amp;gt;g(cos⁡x)⇔sin⁡x&amp;lt;cos⁡xFunction f is strictly decreasing on [0,π4] and strictly increasing on [π4,π2]. As e+1=f(0)=f(π2), for all x∈(0,π2) we have f(x)&amp;lt;e+1

Solving \(\displaystyle{e}^{{{\sin{{x}}}}}+{e}^{{{\cos{{x}}}}}={e}+{1}\)

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