How could I sieve through the extra solutions to $\sin\theta +\cos\theta=\sin2\theta $in the interval $[-\pi,\pi]$?

I am trying to find the number of solutions of the equation $$\sin\theta +\cos\theta=\sin2\theta $$ in the interval $[-\pi,\pi]$.

Here's what I did, $$\sin\theta +\cos\theta=\sin2\theta \\ \Rightarrow (\sin\theta +\cos\theta)^2=(\sin2\theta)^2 \\ \Rightarrow1+\sin2\theta=(\sin2\theta)^2$$

Hence we get $\sin2\theta=\frac{1\pm\sqrt{5}}{2}$, with $\sin2\theta=\frac{1 -\sqrt{5}}{2}$ bein the only valid solution.

Since $2\theta$ is present inside $\sin()$, I assumed that there would be four solutions of the equation. But on plotting the graph I find that there are only two solutions.

Why are there only two solution instead of four, and how could I prevent this mistake in future?

Solution 1:

By only considering $\sin 2\theta=\dfrac{1-\sqrt{5}}{2}$, one gets four roots since $\sin 2\theta$ has a period of $\pi$ and the interval $[-\pi,\pi]$ is twice that length.

But one also has to consider the equality of $\sin \theta + \cos \theta =\dfrac{1-\sqrt{5}}{2}$. Since $$\sin \theta + \cos \theta = \sqrt{2}\sin \left( \frac{\pi}{4} + \theta \right)$$

has a period of $2\pi$, actual number of roots is only $2$.

Solution 2:

If you use the multiple angle formula $\theta=2 \tan^{-1}(x)$ $$\sin(\theta )+\cos(\theta)-\sin(2\theta)=0 \implies x^4-6 x^3+2 x-1=0$$ which has only two real solutions $$x_1=\frac{3+\sqrt{5}}{2}+\sqrt{\frac{1}{2} \left(11+5 \sqrt{5}\right)}$$ $$x_2=\frac{3+\sqrt{5}}{2}-\sqrt{\frac{1}{2} \left(11+5 \sqrt{5}\right)}$$

Then $\theta_1 \sim 2.80847$ and $\theta_2 \sim -1.23768$ that you observed plotting.

Be always careful when you square.

Solution 3:

Avoid squaring whenever practicable as it often introduces

. When do we get extraneous roots?

. Extraneous Roots

et $\sin\theta+\cos\theta=t,t^2=?\le1+1$

$t=t^2-1, 2t=\pm\sqrt5+1$

Now clearly $(\sqrt5+1)^2>2$

$\implies\cos\left(\theta-\dfrac\pi4\right)=\dfrac t{\sqrt2}=?$

As $\cos(2\pi-y)=\cos y$ and cosine has a period $=2\pi,$ we shall have two incongruent solutions

Solution 4:

OP: how can I figure out the correct number of solutions without plotting a graph?

How can I prevent or atleast sieve though the extra solutions? The solutions of this equation itself are too complicated to check manually.

Short of plugging the candidate solutions into the original equation, there is no general method for directly sifting out extraneous ones or determining how many there are.

(If a calculator is allowed, it is typically a good-enough check to just plug in the approximated candidate solutions.)
However, it is possible to catch potential extraneous solutions: carefully watch every line for any step that is not obviously ‘reversible’, noting that every extraneous solution arises from such a step (which is not to say that every such step creates an extraneous solution):
- dropping implicit conditions from trigonometric/logarithmic functions,
- your above squaring step,
- converting absolute value to $\pm$
  
  $\lvert x+6\rvert=2x\\\pm(x+6)=2x\\x=-2 \;\text{ or }\; 6 \text{ (the negative solution is extraneous)}$
- inadvertently multiplying by $0$
  
  $x=1\\x^2=x\\x=0 \;\text{ or }\; 1$
- etc.