Given that the equation of the curve is
[tex]y=3\sin x+2\cos x[/tex]at x=0.
Differentiate y with respect to x to find the slope of the tangent line.
[tex]\frac{dy}{dx}=3\frac{d(\sin x)}{dx}+2\frac{d(\cos x)}{dx}[/tex][tex]\frac{dy}{dx}=3\cos x-2\sin x[/tex]Substitute x=0, we get
[tex]\frac{dy}{dx}=3\cos (0)-2\sin (0)[/tex][tex]\frac{du}{dx}=3[/tex]We get slope =3.
Substitute x=0 in y, we get
[tex]y=3\sin (0)+2\cos (0)[/tex][tex]y=2[/tex]We get the point (0,2).
Recall the formula for the point-slope
[tex]y-y_1=m(x-x_1)[/tex][tex]\text{ Substitute }x_1=0.y_1=2,\text{ and m=3, we get}[/tex][tex]y-2_{}=3(x-0_{})[/tex][tex]y-2_{}=3x[/tex][tex]y_{}=3x+2[/tex]Hence the equation of the line tangent to the curve y=3sinx +2Cosx at x=0 is
[tex]y_{}=3x+2[/tex]