Question

2 sin x-sqrt2=0

Answer 1

Add sqrt2 to both sides, then divide both sides by 2, what do you get?

Answer 2

1?

Answer 3

Nope o_o
If you add sqrt2 to both sides of the equation 2 sin x-sqrt2=0, what do you get?

Answer 4

I don't understand

Answer 5

Do you know to solve y for the equation \(2y - \sqrt{2}=0\) ?

Answer 6

No I don't understand the basics of radians, cos, sin and tan

Answer 7

y is simply a variable. You need not care for radian, sin, cos, or tan.

Answer 8

Alright so I did it with my calculator and I got .71

Answer 9

Ok. Your original question is \(2 sin x-\sqrt2=0\)
Let \(y = sinx\), then the equation becomes \(2y-\sqrt2=0\) and you've just solved y, that is 0.71, which is \(\frac{\sqrt{2}}{2}\).
So, we get \(sinx = \frac{\sqrt{2}}{2}\).
So far so good?

Answer 10

\[2 \sin x - \sqrt{2} = 0\]\[2\sin x = \sqrt{2}\]\[\sin x = \frac{ \sqrt{2} }{ 2 }\]

how you could use the calculator and sin^-1 function (2nd + sin) and get the value, but that answer is not complete. you need to be familiar with radiant circle and find where sinus has value of sqrt(2)/2

at pi/4 and 3pi/4 sinus has the same value, sqrt(2)/2
also, you have to remember that sinus is a periodical function and each of the answers (pi/4 and 3pi/4) you can shift for one or two or three complete circle (2*k*pi, k is the number of circles) and get the same answer

final answers are:
\[x = \frac{ \pi }{ 4 } + 2*k*\pi\]\[x = \frac{ 3*\pi }{ 4 } + 2*k*\pi\]

if you don't understand the basics of radians and trigonometrical functions this will not make much sense to you. you should look for some online sources or textbooks to reiterate that part

Answer 11

yes Callisto I understand so far