# Why does this trig equation have only 2 solutions and not 4?

Upon solving this equation:

$$5\sin x - 5\cos x = 2$$

for the interval $$0 \le x < 360^\circ$$, I manipulated it in order to get a solvable trigonometric equation below:

$$\sin 2x = 21/25$$

Normally this would procure 4 solutions: $$28.6^\circ, 61.4^\circ, 208.6^\circ, 241.4^\circ$$ (one can easily plot this into Desmos to see that it is true).

However, the answers only include the middle 2 solutions: $$61.43^\circ, 208.57^\circ$$. Why is this? Why do 2 of the solutions simply disappear? Could it have something to do with the way in which I manipulated the original equation (by squaring and substituting using the appropriate trigonometric identities)?

After looking at the original equation, it makes sense to some degree, but then I'm not a fan of such methods, especially since one could encounter more complicated equations in the future.

Am I missing/forgetting something fundamental?

• How did you manipulate the original equation to get your $4$ answers ? Jun 2 at 15:06
• You do not need to use a double angle here. Recall that $\cos x=\sin (\frac{\pi}{2}-x)$ and apply difference to product formula. Jun 2 at 15:14

$$\sin{x}-\cos{x}=\frac{2}{5}$$ $$(\sin{x}-\cos{x})^2=\frac{4}{25}$$ $$1-2\sin{x}\cos{x}=\frac{4}{25}$$ $$\sin{2x}=\frac{21}{25}$$
In the second step, when we square both sides, we inadvertently include the solutions of: $$\sin{x}-\cos{x}=-\frac{2}{5}$$ This equation is true for $$28.6^{\circ}$$ and $$241.4^{\circ}$$, which are the extra two solutions which we don't need.