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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 2

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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?

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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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  • $\begingroup$ That's $\cos$, not $\sin$. $\endgroup$ Commented Aug 30, 2015 at 23:57
  • $\begingroup$ Don't you want odd multiples of $\pi/2$? $\endgroup$
    – user84413
    Commented Aug 30, 2015 at 23:57

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