Firstly, let:

4 sin

θ+ 3 cosθ≡Rsin(θ+α)

Then we need:

`R=sqrt(4^2+3^2)=sqrt25=5`

`alpha=arctan(3/4)=36.87^@`

So

4 sin

θ+ 3 cosθ= 5 sin(θ+ 36.87°)

The components of the original function were:

(i) 4 sin *θ* (in black)

(ii) 3 cos *θ* (in blue, with 4 sin *θ*)

When we add these 2 components we get a sine curve that has been shifted to the left by `36.87^@`:

4 sin *θ* + 3 cos *θ* = 5 sin(*θ* + 36.87°) (in red)

From part (a),

4 sin

θ+ 3 cosθ= 5 sin(θ+ 36.87°)

So,

5 sin(

θ+ 36.87°) = 2sin(

θ+ 36.87°) = 0.4

Sine is positive in Quadrants I and II.

Solving sin *α* = 0.4, we get the reference angle *α* = 23.58°.

So the angle for Quadrant I is 23.58° and for Quadrant II, is 180° − 23.58° = 156.42°.

So, we get

`theta+36.87^@=23.58^@`

OR

`theta+36.87^@=156.42^@`

And this gives us:

`theta=23.58^@-36.87^@`

`=-13.29^@`

`=346.71^@`

OR

`theta=156.42^@-36.87^@`

`=119.55^@`

**Are these answers correct?**

We can see from the graph that in the domain 0° ≤ *θ* < 360°, the only two angles which give a value of 2 are 119.6° and 346.7°. So our answer is correct.