How do you solve #6cos^2(x)+7cos(x)+2=0# in the interval [0,360) ??

1 Answer
Douglas K.
Nov 13, 2017

Given: #6cos^2(x)+7cos(x)+2=0#

Please observe that the given equation is a quadratic of the form, #ax^2+bx+c = 0#, where #a =6,b=7,c=2, and x = cos(x)#, therefore, we can use the quadratic formula:

#x = (-b+-sqrt(b^2-4(a)(c)))/(2a)#

Substitute the values for the given equation:

#cos(x) = (-7+-sqrt(7^2-4(6)(2)))/(2(6))#

#cos(x) = (-7+-sqrt(49-48))/12#

#cos(x) = -8/12# and #cos(x) = -6/12#

#x = cos^-1(-2/3)# and #x = cos^-1(-1/2)#

Within the specified domain, #x = cos^-1(-2/3)# has two values:

#x ~~ 131.81^@# and #x ~~ 228.19^@#

Within the specified domain, #x = cos^-1(-1/2)# has two values:

#x = 120^@# and #x = 240^@#

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