Question

2.Solve 2sin (2x) + 2 = 0 on the interval [0, 21).л 3π 9π 11π8' 8' 8' 85л 7л4' 4colaπ 9π857 71 137 1578' 8' 8' 8

Answer 1

,We have

$2\sin (2x)+\sqrt[]{2}=0$

In order to simplify the equation we will substitute the value inside the sine as a new variable

$2x=\theta$

then we have

$2\sin (\theta)+\sqrt[]{2}=0$

then we isolate the variable theta, searching in the unit circle or in a sine graph. In this case we will use the unit circle

As we can see the value of the sin can be taken in the y-axis

$\theta=\sin ^{-1}(-\frac{\sqrt[]{2}}{2})$

we have two values for the interval (0,2pi), that are

$\begin{gathered} \theta=\frac{5\pi}{4} \\ \theta=\frac{7\pi}{4} \end{gathered}$

then we come back to the original variable

for the first solution

$\begin{gathered} 2x=\frac{5\pi}{4} \\ x=\frac{5\pi}{8} \end{gathered}$

for the second solution

$\begin{gathered} 2x=\frac{7\pi}{4} \\ x=\frac{7\pi}{8} \end{gathered}$

then we will calculate the period of the function

$\begin{gathered} 2=\frac{2\pi}{T} \\ T=\frac{2\pi}{2}=\pi \end{gathered}$

then the possible solution can be calculated with the next general formula

$\frac{5\pi}{8}+\pi n$$\frac{7\pi}{8}+\pi n$

therefore using n=0 and n=1 we have

$\frac{5\pi}{8},\frac{7\pi}{8},\frac{13\pi}{8},\frac{15\pi}{8}$

therefore the correct choice is the last option