Question

Please help
(cos(x))/(1+sin(x))=1-sin(x)
Solve the equation
Multiplying both sides by 1+sin(x), using Pythagorean identities, and quadratic formula got me
cos(x)=0 or cos(x)=1
Giving me the specific solutions, on an interval of [0,2pi], 0, pi/2,3pi/2, and 2pi
Yet when I checked it graphically with the original equation 3pi/2 was not an answer
Please help

Answer 1

@jdoe0001 any thoughts?

Answer 2

\(\bf \cfrac{cos(x)}{1+sin(x)}=1-sin(x)\ right?\) checking :)

Answer 3

Yup

Answer 4

And then multiplying by (1+sin(x)) on the RHS is difference of squares so that gets me
1-sin^2(x) which is then cos^2(x)=cos(x)

Answer 5

hmm

Answer 6

I think I see the issue
your solution is correct

however, when taking the inverse trig functions, recall their Range

\(\textit{Inverse Trigonometric Identities}
\\ \quad \\

\begin{array}{cccl}

Function&{\color{brown}{ Domain}}&{\color{blue}{ Range}}\\
\hline\\
{\color{blue}{ y}}=sin^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1&-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\\ \quad \\
{\color{blue}{ y}}=cos^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1& 0 \le {\color{blue}{ y}}\le \pi
\\ \quad \\
{\color{blue}{ y}}=tan^{-1}({\color{brown}{ \theta}})&-\infty\le {\color{brown}{ \theta}} \le +\infty &-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\end{array}\)

notice, the Range for cosine

Answer 7

they all three run on those constraints, thus

Answer 8

Ohh I see
Thanks!