# Thread: Solving Trig Equation - Hidden restriction?

1. ## Solving Trig Equation - Hidden restriction?

Hello everyone,

I arrived at two solutions for the following trigonometric equation, but my textbook's answer only lists one of them.

Could someone please check if I overlooked a restriction or erred during my solving process?

Thank you very much!

---

1. Solve $\displaystyle \sqrt{2} \sin x = \sqrt{3} - \cos x, 0 \leq x \leq 2\pi$.

---

I squared the equation to get:

$\displaystyle 2 \sin^2 x = 3 - 2\sqrt{3}\cos x + \cos^2 x$

Applying Pythagorean Theorem:

$\displaystyle 2(1 - \cos^2 x) = 3 - 2\sqrt{3}\cos x + \cos^2 x$

$\displaystyle -3\cos^2 x + 2\sqrt{3} \cos x - 1 = 0$

Let $\displaystyle y = \cos x$:

$\displaystyle -3y^2 + 2\sqrt{3}y - 1 = 0$

Since y = 0.5773;

therefore: $\displaystyle \cos x = 0.5773$

$\displaystyle x = 0.9554, 5.328$

However, the answer given is only (Scroll with mouse to highlight white text -->) 0.96.

2. Originally Posted by scherz0
Hello everyone,

I arrived at two solutions for the following trigonometric equation, but my textbook's answer only lists one of them.

Could someone please check if I overlooked a restriction or erred during my solving process?

Thank you very much!

---

1. Solve $\displaystyle \sqrt{2} \sin x = \sqrt{3} - \cos x, 0 \leq x \leq 2\pi$.

---

I squared the equation to get:

$\displaystyle 2 \sin^2 x = 3 - 2\sqrt{3}\cos x + \cos^2 x$

Applying Pythagorean Theorem:

$\displaystyle 2(1 - \cos^2 x) = 3 - 2\sqrt{3}\cos x + \cos^2 x$

$\displaystyle -3\cos^2 x + 2\sqrt{3} \cos x - 1 = 0$

Let $\displaystyle y = \cos x$:

$\displaystyle -3y^2 + 2\sqrt{3}y - 1 = 0$

Since y = 0.5773;

therefore: $\displaystyle \cos x = 0.5773$

$\displaystyle x = 0.9554, 5.328$

However, the answer given is only (Scroll with mouse to highlight white text -->) 0.96.
When you square an equation you introduce the possibility of extraneous solutions. All solutions obtained need to be checked by substituting in the original equation.