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

1 Answer
Aug 23, 2017

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