Find the equation of the tangent line at the given value of x

Could someone please help me with this?

Find the equation of a tangent line to f(x)=sin(3x) + cos^2(x)

at x = 0

Thanks

f(x) = sin(3x) + cos^2(x)
f(0) = 0 + 1 = 1
f ' (x) = 3 cos(3x) - 2 cos(x) sin(x)
f ' (0) = 3 - 0 = 3

The equation of tangent with slope 3 and passing through (0, 1) is

y - 1 = 3 (x - 0)
y = 3x + 1

To find the slope of the tangent line, you need to take the derivative of f(x) and plug in 0.

f'(x)= 3cos(3x)-2cos(x)sin(x)
f'(0)= 3

You also need to find the value of f(0).
f(0)= 1

You now have the y-intercept and the slope, which together allow you to write an equation for a line:

y=mx+b

y=3x+1

slope of tangent is df/dx = 3 cos(3x) – 2 sin(x) cos(x) = 3 cos(3x) – sin(2x)
at x = 0, that's 3 cos(0) – sin(0) = 3
and f(0) = sin(0) + cos²(0) = 0 + 1 = 1
so tangent is
y – 1 = 3x ∙ ∙ ∙ y = 3x + 1