Question

If 4sin^2x=1,where 0<x<360 , how many values does x take?

Answer 1

Answer on Question #45769 - Math - Trigonometry

Question.

If $4\sin^2 x = 1$ , where $0 < x < 360{}^\circ$ , how many values does $x$ take?

Solution.

$$4 \sin^ {2} x = 1$$ $$\sin^ {2} x = \frac {1}{4}$$ $$1) \sin x = \frac {1}{2} \text{ or } 2) \sin x = - \frac {1}{2}$$

1) $\sin x = \frac{1}{2}$

$$x = \frac {\pi}{6} + 2 \pi n; \frac {5 \pi}{6} + 2 \pi n; n \in \mathbb{Z}$$

2) $\sin x = -\frac{1}{2}$

$$x = - \frac {\pi}{6} + 2 \pi k; - \frac {5 \pi}{6} + 2 \pi k; k \in \mathbb{Z}$$

We have the solutions in the interval $0 < x < 2\pi$ . Therefore, we use the following values:

1) $x = \frac{\pi}{6}; \frac{5\pi}{6}$

2) $x = \frac{7\pi}{6}; \frac{11\pi}{6}$

See the unit circle to verify (Fig.1):

Fig.1. The unit circle for $\sin \theta$

So, the equation $4 \sin^2 x = 1$ has 4 values:

$$x = \frac {\pi}{6}; \frac {5 \pi}{6}; \frac {7 \pi}{6}; \frac {1 1 \pi}{6}$$

**Answer.**

The equation $4\sin^2 x = 1$ has 4 values:

$$x = \frac {\pi}{6}; \frac {5 \pi}{6}; \frac {7 \pi}{6}; \frac {1 1 \pi}{6} \text{ (i.e. } 30{}^{\circ}; 150{}^{\circ}; 210{}^{\circ}; 330{}^{\circ}).$$

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