Solving an equation with $\mathrm{sin}(x)$ in the exponent: ${2}^{\mathrm{sin}(x)}\cdot \mathrm{cos}(x)+1=1$

Kaiden Wilkins
2022-04-06
Answered

Solving an equation with $\mathrm{sin}(x)$ in the exponent: ${2}^{\mathrm{sin}(x)}\cdot \mathrm{cos}(x)+1=1$

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odvucimo1pp17

Answered 2022-04-07
Author has **23** answers

${2}^{\mathrm{sin}x}\mathrm{cos}x+1=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{2}^{\mathrm{sin}x}\mathrm{cos}x=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{2}^{\mathrm{sin}x}=0\text{or}\mathrm{cos}x=0$

Can you deduce the solution from here?

Can you deduce the solution from here?

Jaime Coleman

Answered 2022-04-08
Author has **3** answers

${2}^{\mathrm{sin}(x)}\mathrm{cos}(x)+1=1$ gives ${2}^{\mathrm{sin}(x)}\mathrm{cos}(x)=0$, so that implies $x=\frac{\pi}{2}+k\pi $ is the only solution, where $k$ is an integer.

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