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
Answer:
Step-by-step explanation:
From
\cos^2x+2\sin x-2=0
Using the identity, we have: \sin^2x+\cos^2x=1 \implying \cos^2x=1-\sin^2x
Opperating:
1-\sin^2x+2\sin x-2=0
-\sin^2x+2\sin x-1=0
\sin^2x-2\sin x+1=0
(\sin x-1)^2=0
\sin x-1=0
\sin x=1
A numerical value for x would be for example x=90 degrees or pi/2 (radians)
And this answer is valid for every angle x=90+360n (n=0,1,2,3,etc) or x=pi/2+2pi*n (n=0,1,2,3,etc)
