for all values of x with 0 ≤ x ≤ 2x such that sinx = cos2x + 1

Question

the answer is π/4, π/2, 5π/4, 3π/2 is there a mathematical way to solve this do I simply just have to plug these in an eliminate?

Answer 1

sin(x)= cos^2(x) +1

sin(x) = 1-sin^2(x) + 1 pythagorean theorem cos^2(x) + sin^2(x) = 1

sin(x) = -sin^2(x) +2 combine constant terms 1+1=2

sin^2(x) +sin(x) -2 = 0

(sin(x)+2)(sin(x) -1) =0 after factoring

sin(x)=1, -2 after setting each factor=0

ignore -2, as sin(x) is always >/= -1

x = pi/2 but NOT 3pi/4, pi/4 or 5pi/4

check the answers

sin(pi/2)=1

cos2(pi/2)+1=0+1=1

sin(pi/4)=sqr2/2

cos2(pi/4)+1=cos(pi/2)+1=0+1=1

sin(3pi/4)= sqr2/2

cos2(3pi/4)+1 = cos(3pi/2)+1 = 0+1 = 1

sin(5pi/4) = -sqr2/2

cos2(5pi/4) +1= cos(5pi/2)+1= 0+1= 1

UNLESS you somehow meant

sinx = cos2x +1, but that seems most unlikely Then

sinx =2cos^2(x)-1+1 using the double angle formula for cosines

sin(x) = 2cos^2(x) combine 1 and -1

sinx= +/-2sqr(1-cos^2(x)) = 2cos^2(x) pythagorean theorem sin^2(x)+cos^2(x) = 1

+/-sqr[1-cos^2(x)] = cos^2(x)

1-cos^2(x)=cos^4(x)

cos^4(x)+cos^2(x)-1= 0

let z=cos^2(x)

z^2 +z -`1 = 0

z=1/2 +/-(1/2)sqr(1+4)

z=.5+/-.5sqr5

x = cos^-1(sqr(.5+/-,5sqr5))

x = some irrational messy angles