# Solving an equation with sin &#x2061;<!-- ⁡ --> ( x ) in the exponent: 2

Solving an equation with $\mathrm{sin}\left(x\right)$ in the exponent: ${2}^{\mathrm{sin}\left(x\right)}\cdot \mathrm{cos}\left(x\right)+1=1$
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Jaime Coleman
${2}^{\mathrm{sin}\left(x\right)}\mathrm{cos}\left(x\right)+1=1$ gives ${2}^{\mathrm{sin}\left(x\right)}\mathrm{cos}\left(x\right)=0$, so that implies $x=\frac{\pi }{2}+k\pi$ is the only solution, where $k$ is an integer.