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`
`sin theta-sqrt2\ cos theta` ≡ `sqrt3\ sin(theta-54.74^@)`
Now, solving the original equation gives:
Sine is positive in Quadrants I and II.
So, from calculator, we get
And this gives us:
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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