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!

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

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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.