Square of equations - Why some solutions rejected

Dear all,

I'm solving 2 sums

a) 1 - cos x = (sqrt 3) sin x

By squaring both sides, and using the sin^2x + cos^2x = 1 relation,

I get 2 cos^2x - cos x - 1=0

Solving, I get x = 120, 240 (for 2nd, 3rd quadrant) and 0, 360 (for 1st, 4th quadrant)

May I know why 240 is rejected?

b) 2 + sin x = 4 cos x

Similarly, I'm solving down to 17 sin^2x + 4 sinx - 12 = 0

I get sin x = 0.7307 or -0.966

Hence, x - 46.9 or 133.1 ; 285 or 255

Why is 133.1 and 255 rejected?

Thanks

Re: Square of equations - Why some solutions rejected

The idea is that $\displaystyle f(x)=g(x)$ implies $\displaystyle (f(x))^2=(g(x))^2$ for all x, but not vice versa. The correct equivalence is

$\displaystyle \text{for all }x,\, [f(x) = g(x)\text{ or }f(x)=-g(x)]\iff (f(x))^2=(g(x))^2$.

Thus, if you solve $\displaystyle (f(x))^2=(g(x))^2$ and find that, say,

$\displaystyle (f(x))^2=(g(x))^2\iff x=x_0\text{ or }x=x_1$,

it does not follow that

$\displaystyle f(x_1)=g(x_1)\text{ and }f(x_2)=g(x_2)$.

At most you can guarantee that

$\displaystyle [f(x_1)=g(x_1)\text{ or }f(x_1)=-g(x_1)]\text{ and }[f(x_2)=g(x_2)\text{ or }f(x_2)=-g(x_2)]$.

Indeed, in your first example

$\displaystyle 1 - \cos(240^\circ)=3/2 = -\sqrt{3}\sin(240^\circ)$.

Re: Square of equations - Why some solutions rejected

Hello, Rodimus79!

;808717]Dear all,

I'm solving 2 sums

Quote:

$\displaystyle a)\;1 - \cos x \:=\: \sqrt 3 \sin x $

By squaring both sides, and using the sin^2x + cos^2x = 1 relation,

I get 2 cos^2x - cos x - 1=0

Solving, I get x = 120^{o}, 240^{o} (for 2nd, 3rd quadrant) and 0^{o}, 360^{o} (for 1st, 4th quadrant)

May I know why 240^{o} is rejected?

When squaring an equation, we __must__ *check our answers*.

Some answers may not check out.

$\displaystyle \text{Check }x = 0^o\!:\;\text{Does }1-\cos 0^o\text{ equal }\sqrt{3}\sin0^o\,?$

. . $\displaystyle \begin{Bmatrix}1-\cos0^o &=& 1-1 &=& 0 \\ \sqrt{3}\sin0^o &=& \sqrt{3}(0) &=& 0 \end{Bmatrix}\;\text{ Yes!}$

$\displaystyle \text{Check }x = 360^o\!:\;\text{Does }1-\cos360^o\text{ equal }\sqrt{3}\sin360^o\,?$

. . $\displaystyle \begin{Bmatrix}1-\cos360^o &=& 1 - 1 &=& 0 \\ \sqrt{3}\sin360^o &=& \sqrt{3}(0) &=& 0 \end{Bmatrix}\;\text{ Yes!}$

$\displaystyle \text{Check }x = 120^o\!:\;\text{Does }1 - \cos120^o\text{ equal }\sqrt{3}\sin120^o\,?$

. . $\displaystyle \begin{Bmatrix}1-\cos120^o &=& 1 - (\text{-}\frac{1}{2}) &=& \frac{3}{2} \\ \sqrt{3}\sin120^o &=& \sqrt{3}(\frac{\sqrt{3}}{2}) &=& \frac{3}{2} \end{Bmatrix}\;\text{ Yes!}$

$\displaystyle \textCheck }x = 240^o\!:\;\text{Does }1-\cos240^o\text{ equal }\sqrt{3}\sin240^o\,?$

. . $\displaystyle \begin{Bmatrix}1-\cos240^o &=& 1-(\text{-}\frac{1}{2}) &=& \frac{3}{2} \\ \sqrt{3}\sin240^o &=& \sqrt{3}(\text{-}\frac{\sqrt{3}}{2}) &=& \text{-}\frac{3}{2} \end{Bmatrix}\;\;{\color{blue}No!}$

Re: Square of equations - Why some solutions rejected

Hello again!

Do you know WHY squaring an equation can introduce "extra" solutions?

Consider the equation: .$\displaystyle 2x \:=\:6$

The answer is, of course: .$\displaystyle x \,=\,3$

Suppose we square the equation before solving.

Then we have: .$\displaystyle (2x)^2 \:=\:6^2 \quad\Rightarrow\quad 4x^2 \:=\:36$

This is a quadratic equation; it will have two solutions.

. . . $\displaystyle x^2 \:=\:9 \quad\Rightarrow\quad x \:=\:\pm3$

We find that $\displaystyle x = 3$ satisfies the equation,

. . but $\displaystyle x = \text{-}3$ does not.

Therefore, $\displaystyle x=\text{-}3$ is an *extraneous* solution.