SOLUTION: Solve for x in the equation (1 +sinx/cosx)+(cosx/1+sinx)=4 for 0≤ x ≤ 2π rad.

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Question 197454: Solve for x in the equation (1 +sinx/cosx)+(cosx/1+sinx)=4 for 0≤ x ≤ 2π rad.
Found 2 solutions by HyperBrain, ikleyn:
Answer by HyperBrain(694) About Me  (Show Source):
You can put this solution on YOUR website!
%281+%2Bsinx%29%2Fcosx+%2B+cosx%2F%281%2Bsinx%29=4
Multiply both sides by cos x (1+ sin x)
%281+%2B+sin+x%29%5E2+%2B+cos%5E2+%28x%29=4cos+x+%281%2B+sin+x%29
1%2B+2+sin+x+%2B+sin%5E2%28x%29+%2B+cos%5E2+%28x%29=4cos+x+%281%2B+sin+x%29
2%2B2sin+x=4cos+x%281%2Bsin+x%29
1%2B+sin+x=2cos+x%281%2Bsin+x%29
Divide both sides by 1 + sin x
2+cos+x=1
cos+x=1%2F2
There are only two solutions:
x=pi/6 or 11pi/6

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve for x in the equation (1 +sinx/cosx)+(cosx/1+sinx)=4 for 0≤ x ≤ 2π rad.
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        The answers x = pi/6 or 11pi/6 in the post by HyperBrain both are incorrect.
        I came to bring a correct solution.


%281+%2Bsinx%29%2Fcosx+%2B+cosx%2F%281%2Bsinx%29=4
Multiply both sides by cos x (1+ sin x)
%281+%2B+sin+x%29%5E2+%2B+cos%5E2+%28x%29=4cos+x+%281%2B+sin+x%29
1%2B+2+sin+x+%2B+sin%5E2%28x%29+%2B+cos%5E2+%28x%29=4cos+x+%281%2B+sin+x%29
2%2B2sin+x=4cos+x%281%2Bsin+x%29
1%2B+sin+x=2cos+x%281%2Bsin+x%29
Divide both sides by 1 + sin x
2+cos+x=1
cos+x=1%2F2
There are only two solutions:
x=pi/3 or 5pi/3

Solved correctly.