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torisob [31]
3 years ago
6

Cos (x + π/ 3) + cos (x - π/ 3) = 1 Solve for "x" on the interval [0, 2pi)

Mathematics
1 answer:
kenny6666 [7]3 years ago
8 0
\bf cos\left( x+\frac{\pi }{3} \right)+cos\left( x-\frac{\pi }{3} \right)=1\\\\
-----------------------------\\\\\
[cos(x)cos\left( \frac{\pi }{3} \right)\underline{-sin(x)sin\left( \frac{\pi }{3} \right)}] 
+ [cos(x)cos\left( \frac{\pi }{3} \right)\underline{+sin(x)sin\left( \frac{\pi }{3} \right)}]=1
\\\\\\

\bf cos(x)cos\left( \frac{\pi }{3} \right)+cos(x)cos\left( \frac{\pi }{3} \right)=1\implies 2cos(x)cos\left( \frac{\pi }{3} \right)=1
\\\\\\
2cos(x)\cdot \cfrac{1}{2}=1\implies cos(x)=1\implies \measuredangle x=cos^{-1}(1)
\\\\\\
\measuredangle x = 
\begin{cases}
0\\
2\pi 
\end{cases}
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____________________________
Explanation:
________________________________________
46 + (x - 3) + (y - 3) = 180 .

46 + 1(x - 3) + 1(y-3) = 180 .

46 + 1x - 3 + 1y - 3 = 180 .

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Subtract "40" from EACH SIDE of the equation:
______________________________________
 40 + x + y - 40 = 180 - 40 ;

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x + y = 140 ;
_____________________________________
Now:
_____________________________________
65 = (x - 3) ;


↔  x - 3 = 65 ;

Add "3" to EACH SIDE of the equation;

    x - 3 + 3 = 65 + 3 ;

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           x = 68 .
______________________________
Now:

Since:  "x + y = 140" ;

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__________________________________
   x + y = 140 ;
 
  68 + y = 140 ;

↔ y + 68 = 140 ;

Subtract "68" from EACH SIDE of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;
______________________________________________
   y + 68 - 68 = 140 = 68 ;

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______________________________________________
So,  solve for "x" and "y".

x = 68, y = 72 .
_______________________________________________
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