Find to the nearest degree, all values of x in the interval 0°≤x

Question

Find to the nearest degree, all values of x in the interval 0°≤x<360° that satisfy the equation 3+tan²x=5tanx.

campbell_st · Answer

well rewrite the equation as 

$$\tan^2(x) -5\tan(x) + 3 = 0$$

you have a quadratic equation, and looking at the coefficients you'll need the 
general quadratic formula to solve for tan(x) 

does that make sense

anonymous · Answer

Yeah that makes sense.. After that, I got 5+/-√12, and simplified the radical, then put them into two separate equations, 5+2√3/2 and 5-2√3/2... Now, what do I do?

campbell_st · Answer

ok... you now need to find the inverse tan values 

looking at 

$$\frac{5 + 2\sqrt{3}}{2}$$

this is a positive angle so it exists in the 1st and 3rd quadrants.... 


so find

$$x = \tan^{-1}( \frac{5 + 2 \sqrt{3}}{2})$$

1st quadrant will be the answer for x

3rd quadrant will be 180 + x

you need to repeat the process for the other value

campbell_st · Answer

you will end up with 4 angles.... 2 in the 1st quadrant and 2 in the 2nd quadrant 

as you need to find

$$x = \tan^{-1}(4.23)...... and..... x = \tan^{-1}(0.7679)$$

anonymous · Answer

But first, do I put the whole tan inverse in the calculator like this: $$\tan^{-1}$$(5+2√3/2) or $$\tan^{-1}$$(5+2√3)/2 ? Does that make sense?

campbell_st · Answer

oops should be 2 angles in the 3rd quadrant

campbell_st · Answer

yep thats it.... just as easy to make it a decimal value.... since you are using inverse tan.... to find angles

anonymous · Answer

Thank you so much, I found the answers to the problem, you're a major help!! Again, thank you!

campbell_st · Answer

well done