How do you find a numerical value of one trigonometric function of x given #cos^2x+2sinx-2=0#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Nov 14, 2016 #:.x=pi/2 + 2pin# Explanation: #cos^2x+2sinx-2=0# #(1-sin^2x)+2sinx-2=0# #-sin^2x+2sinx-1=0# #sin^2x-2sinx+1=0# Now factor. #(sinx-1)(sinx-1)=0# #(sinx-1)^2=0# #sinx-1=0# #sinx=1# #x=sin^-1 1# #:.x=pi/2 + 2pin# Answer link Related questions What does it mean to prove a trigonometric identity? How do you prove #\csc \theta \times \tan \theta = \sec \theta#? How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#? How do you prove that #(2sinx)/[secx(cos4x-sin4x)]=tan2x#? How do you verify the identity: #-cotx =(sin3x+sinx)/(cos3x-cosx)#? How do you prove that #(tanx+cosx)/(1+sinx)=secx#? How do you prove the identity #(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)#? See all questions in Proving Identities Impact of this question 11468 views around the world You can reuse this answer Creative Commons License