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?

1 Answer

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