So, I'm struggling immensely with my precalc homework concerning trig functions. I'm not looking for just answers but an explanation as well, so I can understand in the future.If you can help me, thank you so much. :)
Solve 2cos^2x=√3cosx for principal values of x. Express in degrees.
Solve 2cos^2x=√3cosx for principal values of x. Express in degrees.
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2cos^2x=√3cosx this is 2 times cos²x = √3 times cosx
we can divide both sides by cosx
2 cos²x / cosx = √3 cosx / cosx
2 cosx = √3 now divide by 2
2 cosx /2 = √3/2
cosx = √3/2 if you know your basic angles (30-60) and (45-45) you will know that x can be
x= π/6 or 30° and 11π/6 or 330°
we can divide both sides by cosx
2 cos²x / cosx = √3 cosx / cosx
2 cosx = √3 now divide by 2
2 cosx /2 = √3/2
cosx = √3/2 if you know your basic angles (30-60) and (45-45) you will know that x can be
x= π/6 or 30° and 11π/6 or 330°
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Like with any quadratic equation, you solve by factoring (if feasible);
otherwise fall back to the quadratic formula.
Your 2cos^2x=√3cosx becomes 2(cos² x) -(√3) cos x =0 which
factors: (cos x) [2 cos x -√3] = 0
Either factor can be zero; so if cos x = 0 then x =90°
If the other factor is zero, cos x =(√3)/2 so x = 60°
==========
BTW you should use grouping symbols judiciously so there's
no ambiguity (abt what the radicand is; and what the exponent is);
otherwise fall back to the quadratic formula.
Your 2cos^2x=√3cosx becomes 2(cos² x) -(√3) cos x =0 which
factors: (cos x) [2 cos x -√3] = 0
Either factor can be zero; so if cos x = 0 then x =90°
If the other factor is zero, cos x =(√3)/2 so x = 60°
==========
BTW you should use grouping symbols judiciously so there's
no ambiguity (abt what the radicand is; and what the exponent is);