One way can be using tan\frac x2=t so sin x=\frac{2t}{1+t^2} and cos x=\frac{1-t^2}{1+t^2}. Here 2sin x= cos x implies t^2+4t-1=0 from wich tan \frac x2=2\pm\sqrt{5}.Hence the answer of ...

\displaystyle{x}={0} Explanation: \displaystyle{1}+{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}{\quad\text{or}\quad}{\cos{{x}}}-{\sin{{x}}}={1} . Squaring both sides we get \displaystyle{\left({\cos{{x}}}-{\sin{{x}}}\right)}^{{2}}={1}{\quad\text{or}\quad}{{\cos}^{{2}}{x}}+{{\sin}^{{2}}{x}}-{2}{\sin{{x}}}{\cos{{x}}}={1}{\quad\text{or}\quad}{1}-{\sin{{2}}}{x}={1}{\quad\text{or}\quad}{\sin{{2}}}{x}={0}={\sin{{0}}}; ...

Bdub Nov 12, 2016 Pick a value for x like \displaystyle\frac{\pi}{{3}} and plug it in to both side to show that they don't equal each other and therefore not an identity

As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, ...

We have \tan(x) = 3 Draw a right angle triangle where x is one of the angle, let the opposite side be of length 3, let the adjacent side be of length one. Then by Pythagoras Theorem, the ...

Hint Maybe starting with \sin^2 x + \cos^2 x = 1 is helpful :)