The first part of the question asks me to square both sides of the equation: $$3 \cos θ=2 − \sin θ$$

So that I can obtain and solve the quadratic: $$10\sin^2θ−4\sinθ−5=0 \;\;\text{for}\;\; 0 ≤ θ<360°$$

solutions obtained in the interval are: $69.2°, 110.8°, 212.3°, 327.7°$

However the second part of this question stumps me, it asks:

"Explain why not all of the answers satisfy $3\cosθ=2−\sinθ$."

I understand what it's asking, but I dont get why this is so. Any help in understanding this would be appreciated, thank you.


We risk introducing new "solutions" when we solve an equation by squaring it:

Suppose, for example, we are given the equation, and are asked to find all real solutions: $$x^{3/2} = 2\sqrt x\tag{1}$$

We can solve it by first squaring it: $$(x^{3/2})^2 = (2\sqrt x)^2 $$ $$\iff x^3 -4x = 0 $$ $$\iff x(x^2 - 4) = 0 $$ $$\iff x(x+2)(x-2) = 0\tag{2}$$ $$ \iff x = 0, \;\text{ or} \; x= -2, \text{ or}\;x = 2$$

Three solutions, right?

Wrong: If we go back to the original equation $(1)$, we see the equation is undefined on the reals when $x = -2$, since we cannot take the square root of a negative value. So, while $x = -2$ solves equation $(2)$, it is not a solution to equation $(1)$.

$x = 0,\; x = 2$ are, however solutions to both $(1)$ and $(2)$.

So while squaring each side of an equation often allows us to more easily obtain solutions, we must always be careful to check solutions in the original equation to confirm, or throw out, the solutions to the squared equation.

  • $\begingroup$ Mornin Amyyyyyyy $\endgroup$ – mrs Mar 17 '13 at 4:58
  • $\begingroup$ @Babak: or you, I hope! For me, midnight! ;) $\endgroup$ – Namaste Mar 17 '13 at 5:00
  • $\begingroup$ OMG! Please rest! We need a complete sleep. Please. Have a nice an Butterfly Night. GN Amy. ;-) $\endgroup$ – mrs Mar 17 '13 at 5:03
  • $\begingroup$ Please leave the site alone. Think about yourself. You need power to continue. I will be waiting for you for next hours. $\endgroup$ – mrs Mar 17 '13 at 5:07
  • $\begingroup$ Thanks for clearing up the confusion! $\endgroup$ – seeker Mar 17 '13 at 10:11

Your reasoning is a sequence of logical consequences (and not equivalences) that lead you to conclude that $\theta\in \{69.2, \,110.8, \,212.3, \,327.7\}:=S$.

You don't necessarily know that every $\theta \in S$ will satisfy the initial equation, you just know that every solution to the equation will be in $S$.

In fact it doesn't happen that every element of $S$ will be a solution to the equation. Note that for every $\theta\in S$ we have $2-\sin (\theta)>0$. However you can check that not all $\theta\in S$ will make the inequality $3\cos (\theta)>0$ true.

The reason for all this was that you squared the equation. Note that $2^2=(-2)^2$ and $2\neq -2$.

As an example of a similar problem consider the equation $\cos (\theta)=\sin (\theta)$ in $[0,360]$. You should know that the solutions to this equation are $45º$ and $225º$. However if you proceed in a similar way you get

$$\cos (\theta)=\sin (\theta) \Longrightarrow (\cos (\theta))^2=(\sin (\theta))^2 \Longrightarrow \theta \in \{45º, 135º, 225º, 315º\}$$

You can "go forward" in your reasoning, but you can't "go backwards".


There is also the purely trigonometric approach: $$ 3\cos\theta= 2-\sin\theta\quad\Leftrightarrow\quad 3\cos\theta+\sin\theta=2\quad \Leftrightarrow\quad \frac{3}{\sqrt{10}}\cos\theta+\frac{1}{\sqrt{10}}\sin\theta=\frac{2}{\sqrt{10}} $$ Now setting $\theta_0:=\arcsin (3/\sqrt{10})$ and $\theta_1:=\arcsin(2/\sqrt{10})$, this is equivalent to $$ \sin(\theta+\theta_0)=\sin\theta_1. $$ So the solutions are $$ \theta_1-\theta_0+2\pi\mathbb{Z}\quad\mbox{and}\quad \pi-\theta_1-\theta_0+2\pi\mathbb{Z}. $$ In $[0,2\pi]$, this leaves $$ \theta_1-\theta_0+2\pi\simeq 5,72\;\mbox{radians}\quad\mbox{and}\quad \pi-\theta_1-\theta_0\simeq 1.20 \;\mbox{radians} $$ hence $$ \simeq 327,7\;\mbox{degrees}\quad \mbox{and}\quad \simeq 68,7\;\mbox{degrees}. $$


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