1 Answer

Jjkparker · Answer 1 · 2023-06-05T20:20:20+0000

,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

$\begin{gathered} \theta=(5\pi)/(4) \\ \theta=(7\pi)/(4) \end{gathered}$

then we come back to the original variable

for the first solution

$\begin{gathered} 2x=(5\pi)/(4) \\ x=(5\pi)/(8) \end{gathered}$

for the second solution

$\begin{gathered} 2x=(7\pi)/(4) \\ x=(7\pi)/(8) \end{gathered}$

then we will calculate the period of the function

$\begin{gathered} 2=(2\pi)/(T) \\ T=(2\pi)/(2)=\pi \end{gathered}$

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

$(5\pi)/(8)+\pi n$
$(7\pi)/(8)+\pi n$

therefore using n=0 and n=1 we have

$(5\pi)/(8),(7\pi)/(8),(13\pi)/(8),(15\pi)/(8)$

therefore the correct choice is the last option

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