Firstly, express the LHS in the form R sin(θα) (Note the negative sign!):

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

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



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

Now, solving the original equation gives:

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


Sine is positive in Quadrants I and II.

So, from calculator, we get




And this gives us:




Are these answers correct?

Check graph trig equation

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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