Question

Solve the equation for exact solutions over the interval [0,
2π).



4sin2x+8sinx+4=0

Answer 1

Answer on Question #49861 – Math – Trigonometry

Question:

Solve the equation for exact solutions over the interval $[0, 2\pi)$. $4\sin^2 x + 8\sin x + 4 = 0$.

Solution:

Let $\sin x = t$, then we obtain the equation:

$$4t^2 + 8t + 4 = 0,$$

This equation has one root:

$$D = b^2 - 4ac = 8^2 - 4 \cdot 4 \cdot 4 = 0,$$ $$t = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8}{2 \cdot 8} = -1.$$

So, let’s back to the substitution $\sin x = t$:

$$\sin x = -1.$$ $$x = \frac{3\pi}{2} + 2\pi n, \, n \in \mathbb{Z}$$

So, we can see that in the interval $[0, 2\pi)$ the equation $4\sin^2 x + 8\sin x + 4 = 0$ has one solution:

$$x = \frac{3\pi}{2}.$$

Answer:

$$x = \frac{3\pi}{2}.$$

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