How do you solve $sinx + cosx = 1$ over the interval 0 to 2pi?

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How do you solve $sinx + cosx = 1$ over the interval 0 to 2pi?

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Answer 1 · 2016-03-09T14:22:21

0, $pi/2$ and $2pi$ ..

Explanation:

With t = tan $(x/2)$ , use sin x = $(2t)/(1+t^2) and cos x = (1-t^2)/(1+t^2)$ .
Then, t $(t-1)$ = 0. t = tan $(x/2)$ = 0 gives x = 0, 2 $pi$ .
t = tan $(x/2)$ = 1 gives x = $pi/2$ ..

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How do you solve $sinx + cosx = 1$ over the interval 0 to 2pi?

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How do you solve $sinx + cosx = 1$ over the interval 0 to 2pi?