So I'm looking at a worked example in my textbook and it's showing me how to find the x-intercepts when sketching a trig graph. For example

y= sin (x + pi/4)

0= sin (x + pi/4)

So x + pi/4 = 0, pi, 2pi ----> x= -pi/4, 3pi/4, 7pi/4

My question is, where did the 0, pi and 2pi come from?

I tried doing the same approach for this question

y= 2sin2(x-pi/2) - 2 but it's not working out.

Instead, I did this

0 = 2sin 2(x-pi/2) - 2

1 = sin 2(x-pi/2)

sin^-1 (1) = 2(x-pi/2)

pi/2 = 2(x-pi/2)

Sin is positive in quadrants 1 and 2

2(x-pi/2) = pi/2, pi/2 + 2pi = 5pi/2

(x-pi/2) = pi/4, 5pi/4

x = 3pi/4 , 7pi/4

Could someone explain what I did differently from what the worked example did?

I feel like my way is alot more time consuming than the worked examples....

Sorry this is a mess!

Thankyou!

## Finding x-intercepts when sketching trigonometry graphs

Answered## 1 Answer