Find all real numbers in the interval [0,2pi) that satisfy the equation

Because cos² + sin² = 1 you can turn this into a quadratic for sin by adding sin²x to both sides:

 

23sin²x - 2sin(x) + sin²x = cos²x + sin²x

so

24sin²x - 2sin(x) = 1

or 

24sin²x - 2sin(x) - 1 = 0

 

This can be solved for sinx:  

sinx = 1/4  and sinx = -1/6

 

Now find sin^-1x

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)} = {\mathtt{14.477\: \!512\: \!185\: \!93^{\circ}}}$$  =14.5° to nearest tenth of a degree (you will have to turn this into radians if that is what is wanted)

 

You can do the same for the other value.

 

Since it asks for all angles between 0 and 2pi that satisfy the equation, you should think about where else sinx = 1/4 etc.  (hint: look in the second quadrant when sinx is positive and in the third and fourth quadrants when sinx is negative).

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