Firstly, express the LHS in the form *R* sin(*θ* −
*α*) (Note
the negative sign!):

`sin theta-sqrt2 cos theta` ≡ `R sin(theta-alpha)`

`R=sqrt(1^2+(sqrt2)^2)=` `sqrt(1+2)=` `sqrt3`

`alpha=arctan(sqrt2/1)=54.74^@`

So

`sin theta-sqrt2\ cos theta` ≡ `sqrt3\ sin(theta-54.74^@)`

Now, solving the original equation gives:

`sqrt3\ sin(theta-54.74^@)=0.8`

`sin(theta-54.74^@)=0.8/sqrt3=0.4619`

Sine is positive in Quadrants I and II.

So, from calculator, we get

`theta-54.74^@=27.51^@`

OR

`theta-54.74^@=152.49^@`

And this gives us:

`theta=27.51^@+54.74^@=82.25^@`

OR

`theta=152.49^@+54.74^@=207.23^@`

**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 0.8 (the red dotted line) are `82.3^@` and `207.2^@`. So our answer is correct.

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