Solving trigonometric system of equations

Question

What are the solutions for this system of equations when $\alpha \in \mathbb{R}$ is considered a constant and $0 \leq x < 2\pi$. $$ I) \ (y - \cos x)\sin x + (\alpha - \sin x) (-\cos x) = 0$$ $$ II) \ (y - \cos x) = 0$$

My attempt:

From II) we know that $y = \cos x$.

If we apply that in I), we get: $$(\alpha - \sin x) (-\cos x) = 0 \Leftrightarrow$$ $$- \cos x = 0 \vee \alpha = \sin x$$

Now the first pair of solutions is $(\frac{\pi}{2},0),(\frac{3\pi}{2},0)$. However I fail to express the other solutions in terms of $\alpha$. Any help would be very much appreciated.

Answer 1

$$ \begin{cases} (y−\cos{x})\sin{x}+(α−\sin{x})(−\cos{x}) = 0 \\ y − \cos{⁡x} = 0 \end{cases} \tag{1} $$ First, substitute all instances of $(y−\cos{x})$ with $0$. $$ \begin{cases} \color{red}{(0)}\sin{x}+(α−\sin{x})(−\cos{x}) = 0 \\ \color{red}{(0)} = 0 \end{cases} \tag{2} $$ Now, we are left with one equation. $$ (α−\sin{x})(−\cos{x}) = 0 \tag{3} $$ We need to find the zeros of each term.

$$ α−\sin{x} = 0 \tag{4} $$

$$ -\sin{x} = -a \tag{4.1} $$

$$ \sin{x} = a \tag{4.2} $$

$$ x = \arcsin{a} \\ \operatorname{or} \\ x = \pi - \arcsin{a} \tag{4.3} $$

We have found one set of $x$ values, in terms of $a$. Next, we zero the second term.

$$ −\cos{x} = 0 \tag{5} $$

$$ \cos{x} = 0 \tag{5.1} $$

$$ x = \pm\arccos{0} \tag{5.2} $$

$$ x = \pm\frac{\pi}{2} \tag{5.3} $$

$-\pi/2$ is outside our interval of $x$, $[0,2\pi)$, but $-\frac{\pi}{2}$ is coterminal with $\frac{3\pi}{2}$; thus

$$ x = \frac{\pi}{2} \\ \operatorname{or} \\ x = \frac{3\pi}{2} \tag{5.4} $$

Finally, we have found all values of $x$.

$$ x = \{ \arcsin{a} , \pi - \arcsin{a} , \frac{\pi}{2} , \frac{3\pi}{2} \} \tag{6} $$

Next, solve the second equation in the original system for $y$.

$$ y = \cos{⁡x} \tag{7} $$

Then, substitute each value of $x$ to find the corresponding value of $y$.

$$ y = \cos{( \arcsin{a} )⁡} \ | \ x = \arcsin{a} \tag{8} $$

$$ y = \sqrt{ 1 - a^2 ⁡} \ | \ x = \arcsin{a} \tag{8.1} $$

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$$ y = \cos{( \pi - \arcsin{a} )⁡} \ | \ x = \pi - \arcsin{a} \tag{9} $$

$$ y = -\sqrt{ 1 - a^2 ⁡} \ | \ x = \pi - \arcsin{a} \tag{9.1} $$

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$$ y = \cos{ \frac{\pi}{2} ⁡} \ | \ x = \frac{\pi}{2} \tag{10} $$

$$ y = 0 \ | \ x = \frac{\pi}{2} \tag{10.1} $$

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$$ y = \cos{ \frac{3\pi}{2} ⁡} \ | \ x = \frac{3\pi}{2} \tag{11} $$

$$ y = 0 \ | \ x = \frac{3\pi}{2} \tag{11.1} $$

In conclusion, the values of $x$ and $y$ that satisfy the system of equations are $$ ( \arcsin{a} , \sqrt{ 1 - a^2 ⁡} ) , \\ ( \pi - \arcsin{a} , -\sqrt{ 1 - a^2 ⁡} ) , \\ ( \frac{\pi}{2} , 0 ) , \\ ( \frac{3\pi}{2} , 0 ) $$

Also, note that the system is only solvable when $-1 \le a \le 1$. Depending on the value of $a$, the $x$ value of the first point may fall outside the interval, so a coterminal value of $x$ may be required.

Answer 2

$$ I) \ (y - \cos x)\sin x + (\alpha - \sin x) (-\cos x) = 0$$ $$ II) \ (y - \cos x) = 0$$

To solve for two variables x and y, you need two equations, that much is right.

$ \alpha $ is a given constant say 1/2, is given, it is not to be solved for !

As you say after plugging in $ II)$ into $ I): $ $(\alpha - \sin x) (-\cos x) = 0$

which has two sets of solutions one for $ x = \pi/6, 5 \pi/6,...$ and another, $ \pi/2, 3 \pi/2,$ ...with co-terminal angles.

The corresponding set of solutions for $ y = ( \sqrt 3/2 , - \sqrt 3/2, ..),(0,0,....)$

EDIT 1:

$ x = \sin^{-1}\alpha,\, \pi- \sin^{-1}\alpha..$ and another $ \pi/2, 3 \pi/2,$ ...with co-terminal angles.

set of solutions for $ y = ( \pm \cos^{-1}\alpha),(0,0,....). $