# $2\sin x=3\cot x$ 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.

Going from your 3rd line

$$2\sin^2 x = 3\sqrt{1-\sin^2 x}$$

to your 4th line

$$\left(\frac{2\sin^2}{3}\right)^2 = 1 - \sin^2 x$$

you have taken the square of both sides of the equation. However, taking squares is dangerous, because you're potentially introducing spurious solutions when you take square roots later -- going from $y^2$ to $\pm y$ you have to test both solutions to see which one(s) work.

As a very simple motivating example, consider

$$x=3$$

You can square both sides to get

$$x^2=9$$

but when you take the square root and write

$$x=\pm3$$

only one possibility satisfies the original equation.

• Well that does make sense. But how do you look out for this sort of stuff in say like an exam – Baber Feb 19 '14 at 12:57
• @Baber Two ways: (i) it's good practise when you're solving an equation to check your answers by substituting into the original equation; (ii) when you square, make a note somewhere to check you haven't got spurious solutions. – TooTone Feb 19 '14 at 13:00

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.