Part (a)

Firstly, let:

4 sin θ + 3 cos θR sin(θ + α)

Then we need:

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

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

So

4 sin θ + 3 cos θ = 5 sin(θ + 36.87°)

What have we done?

The components of the original function were:

(i) 4 sin θ (in black)

4 sin theta

(ii) 3 cos θ (in blue, with 4 sin θ)

3 cos theta

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)

5 sin (theta + 36)

Part (b)

From part (a),

4 sin θ + 3 cos θ = 5 sin(θ + 36.87°)

So,

5 sin(θ + 36.87°) = 2

sin(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?

Graph trig equation solution

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.