finding x intercepts of sin graph

cos(x) = 0   at 90, 270, 450, ....   90+n*180    an infinite number of x intercepts.

Since you have a 2*x term,  you will have intercepts every 90 degrees.

But what is your function.    f(x) = 2*cos( 2*x -1)   ???

The first intercept for that is       2*x -1 = 90  ==>  x = 91/2 = 45.5

sorry you will need to explain this more. my function is (x) = 2*cos( 2*x)  -1
 the first intercept is simply 30 deg, and to find the other intercepts what mathematics are we doing ?

if sqrt(2) *sin( x) + 1 there are only 2 intecepts in firs 2 pie. I can find one OK but the 2nd ?

Cosine is 0 at   ... -90, 90, 270, 450..... , 90+n*180,...      where n is any integer

2cos (2x-1)=0

gives

  2x-1 = 90+n*180              n any integer

  x= (91 + n*180)/2    

  x =45.5 + n*90

Different values of n give different solutions,

for example  n= -2, -1, 0, 1, 2.. we get

x = -134.5,  -44.5,  45.5, 135.5,  225.5

2*cos( 2*x)  -1 = 0 is where it intercepts.
So 2*cos( 2*x) = 1
cos( 2*x) = 1/2

So you know that 30 degrees is the first intercept. The others will be anywhere else cos(2*x) = 1/2

Where does cos = 1/2? In one cycle, there are two places, then every intercept is 360n + one of them (where n is any integer).

For example, cos(60) = 1/2, cos(120) = 1/2 so cos (360*4 + 60) = 1/2 and cos (360*4 + 120) = 1/2

Of course you have 2x inside the cos, so you need to do some more math there.

Yeah. Oops. It's 60 and -60, not 60 and 120. Replace everywhere I said 120 with either -60 or 300 whichever you prefer.

..and a whoops for me    360 - 2x = 60    => x = 210    not 150

No. 150 was right.

TommySzalapski.  Ty, I think I am loosing my marbles today.

all this about every 90 degrees gets to the bottom on the original question
BUT there is a big difficulty with finding the first intercept.
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"cos2x=.5=60 deg
x=30deg is 1 intercept but what about 3 others ? "
What is this all about.  I am lost
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" 2*x -1 = 90"
you are mixing degrees and radians
stick to radians