SOLUTION: Solve for x. (sin x + cos x)/(1 - tan x) = 0

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Question 1208813: Solve for x.

(sin x + cos x)/(1 - tan x) = 0

Found 2 solutions by AnlytcPhil, ikleyn:
Answer by AnlytcPhil(1810) About Me  (Show Source):
You can put this solution on YOUR website!

%28sin%28x%29+%2B+cos%28x%29%29%2F%281+-+tan%28x%29%29+=+0, tan%28x%29%3C%3E1, x%3C%3E45%5Eo%2B360%5Eo%2An

sin%28x%29%2Bcos%28x%29=0 

sin%28x%29=-cos%28x%29

sin%28x%29%2Fcos%28x%29=-1, cos%28x%29%3C%3E0, x%3C%3E90%5Eo%2B360%5Eo%2An

tan%28x%29=-1

x=315%5Eo%2B360%5Eo%2An

Edwin




Answer by ikleyn(52915) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve for x: (sin x + cos x)/(1 - tan x) = 0
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        The solution in the post by Edwin is incomplete.
        I came to bring a complete solution.


The domain (the set of real numbers, where left side of the equation
is defined) is  tan(x) =/= 1.

So, the prohibited values of x are  pi%2F4+%2B+k%2Api,  k = 0, _/-1, +/-2, . . . 


We are looking for solutions of the given equation that are in its domain.


In the domain, the given equation is equivalent to

    sin(x) + cos(x) = 0,    (1)

or

    sin(x) = -cos(x).       (2)


The solutions to this equation can not be with cos(x) = 0 (since then sin(x) = 1, 
and this equation is not held) .


Therefore, we can divide both sides of equation (2) by cos(x).  We get then

    tan(x) = -1.


The solutions to this equation are

    x = %283%2F4%29%2Api+%2B+k%2Api,  or  135° + 180°*k,  k = 0, +/-1, +/-2, . . . 


ANSWER.  The solutions to the given equation are   x = %283%2F4%29%2Api+%2B+k%2Api,  or  135° + 180°*k,  k = 0, +/-1, +/-2, . . . 

Solved.