Determine all points of intersection: y=cosx and y=sinx in the first quadrant

So the question is asking to determine all points of intersection, and base on my research and further computation about the said question, the possible answer would be y=cos(x), sin(x)=cos(x),  sin2(x)=cos2(x), sin2(x)=1−sin2(x),  sin2(x)+sin2(x)=1,  2sin2(x)=1, sin2(x)=1/2, sin(x)=√1/2, sin(x)=√2/2. I hope this would help 

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

W-14=t. that i think is the answer idk