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July 29, 2014

July 29, 2014

Posted by **Kelly** on Monday, March 14, 2011 at 6:38pm.

2cos(x)-3=0

- Trig -
**bobpursley**, Monday, March 14, 2011 at 6:40pmIs this cosx=1/2sqrt3 ?

- Trig -
**Kelly**, Monday, March 14, 2011 at 6:42pmno it's 2cosx-sqrt3=0

- Trig -
**Reiny**, Monday, March 14, 2011 at 6:44pmI will read it as

2cosx - √3 = 0

cosx = √3/2

x = 30 , from the standard 30-60-90 triangle

or x = 330

"all solutions " would be

x = 30 + k(360) or x = 330+k(360) where k is an integer.

- Trig -
**Kelly**, Monday, March 14, 2011 at 7:06pmHow do you find the solutions?

- Trig -
**Reiny**, Monday, March 14, 2011 at 7:34pmsince the standard cosine curve has a period of 360, any answer you have will repeat itself in the next curve or 360 either to the right or to the left.

by multiplying 360 by k, k and integer, I am simply adding/or subtracting multiples of 360 to any answer.

to get my original 30, as I said, I recognized the rations of the 30-60-90 right-angled triangle, which are 1:√3:2.

Secondly , since the cosine is positive in the 1st and 4th quadrant, the 30 will be the 1st quadrant solution and 360-30 or 330 would be the 4th quadrant solution.

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