Question

How do you solve `2sin^2x-1=0`?

Answer 1

45 and 135

Explanation:

#2 sin^2(x)-1=0

sin^2(x)=1/2

take `sqrt(sin(x)^2)=sqrt(1/2)`

`sin(x)=sqrt2/2`

sin = opposite / hypotenuse

So, lets think about a right triangle with opposite `sqrt2` and hypotenuse `2`

and get the adjacent by Pythagorean theorem `sqrt(2^2-(sqrt2)^2`

`=sqrt2`

So, we have an isosceles right triangle so

we have right angle = 90

and two equal angles equal 90

so one angle equal to 45

and we have that Unit circle gives another arc (angle), that has the same sin value

x=180-45= 135