solutions of $\sin(x)+\cos(x)-3\cdot\sin(x)\cos(x)+1=0$

Question

I need to find the solutions of $$\sin(x)+\cos(x)-3\cdot\sin(x)\cos(x)+1=0$$

My try: I rewrite the equation: $\sin(x)+\cos(x)-\frac{3}{2}\cdot\sin(2x)+1=0\Rightarrow \frac{3}{2}\sin(2x)-1=\cos(x)+\sin(x)$ then I squared both sides and I got $\frac{9}{4}\sin^{2}(2x)-3\sin(2x)-\sin(2x)=0$

I noted $\sin^{2}(2x)=t$ and I got $t=0; t=16/9$ so $\sin(2x)=0$ and I got $x\in\left \{ \frac{k\pi}{2}|k\in\mathbb{Z} \right \}$ but the right answer is $(2k+1)\pi|k\in Z$

Where's my mistake?

Answer 1

Hint

Avoid squaring which often introduces extraneous root

Extraneous Roots

Set $a=\sin x+\cos x,a^2=1+?$

Observe that for real $x,a^2\le2$

Answer 2

If you take $\sin {2x} = t $, then you get $\frac{9}{4} t^2 - 4 t = 0 \implies t= 0 , t = \frac{16}{9}$. If you take $\sin^2 {2x} = t $, then you get $\frac{9}{4} t - 4 \sqrt{t} = 0 \implies t= 0 , t = (\frac{16}{9})^2$.

Now since $-1\le \sin x \le 1$, so $\sin 2x = 0$.

Hence $x=\frac{n\pi}{2}$, for all $n \in \mathbb{Z} $.

But the equation $\sin 2x=0$ also satisfy the equation $x=(2n+1)\pi$ for $n \in \mathbb{Z} $. So if the option contains only $x=(2n+1)\pi$ for $n \in \mathbb{Z} $, then you have to choose it as correct option.

Answer 3

There's a general approach if you are wondering where the transformations come from.

Replace $\theta$ with $(\theta'-\theta_0)$. Then find a $\theta_0$ such that the product of sines and cosines in $\theta'$ vanishes . This eliminates the cross terms allowing you to complete the squares for sine and cosine.

In this case, the right selection of $\theta_0$ well not only eliminate the cross term, but the term linear in sine as well leaving a quadratic equation in terms of $\cos(\theta')$.