Find a numerical value of one trigonometric function of x for cos^2x+2sinx-2=0
Accepted Solution
A:
Answer:[tex]\sin x=1[/tex]Step-by-step explanation:The given function is [tex]\cos^2x+2\sin x-2=0[/tex]We use the identity: [tex]\sin^2x+\cos^2x=1[/tex] [tex]\implies \cos^2x=1-\sin^2x[/tex] This implies that:[tex]1-\sin^2x+2\sin x-2=0[/tex][tex]-\sin^2x+2\sin x-1=0[/tex][tex]\sin^2x-2\sin x+1=0[/tex][tex](\sin x-1)^2=0[/tex][tex]\sin x-1=0[/tex][tex]\sin x=1[/tex]Hence the numerical value of one trigonometric function(the sine function) is 1