Hi I need help with a trig problem:
I have $2^{\sin(x)} \cdot \cos(x) + 1$, and I need this to equal $1$ between $x = -3$ and $3$. I keep going in circles with substitution, etc. Any help would be greatly appreciated.
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2 Answers
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$2^{\sin x } \cos x +1 = 1 \implies 2^{\sin x } \cos x = 0 \implies 2^{\sin x}=0 \text{ or } \cos x =0$.
Can you deduce the solution from here?
answered Aug 30, 2015 at 23:48
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$2^{\sin(x)}\cos(x)+1 = 1$ gives $2^{\sin (x)}\cos(x)=0$, so that implies $x=\frac{\pi}{2} + k\pi $ is the only solution, where $k$ is an integer.
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$\begingroup$ Don't you want odd multiples of $\pi/2$? $\endgroup$ Aug 30, 2015 at 23:57