<s>[INST] $2\sin x=3\cot x$ from $0^\circ$ to $360^\circ$ I used two slightly different approaches to solve this.  First approach gives 2 correct solutions,  second approach gives 4 solutions of which 2 are correct and 2 wrong,  I just cannot figure out why I'm getting 2 extra wrong answers with second approach. 


 [/INST] When you write $$2\sin^2 x=3\sqrt{1-\sin^2 x}$$ $$\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x$$ you're really saying $$2\sin^2 x=3\sqrt{1-\sin^2 x}\Rightarrow\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x$$ It's false that $$\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x\Rightarrow 2\sin^2 x=3\sqrt{1-\sin^2 x}$$ because $f(x)=x^2$ isn't an injective function. From that, you get that all the solutions to the first equation are solutions to the last equation, but not all the solutions to the last equation are solutions to the first equation.
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