Question #a070e

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Answer 1 · 2017-02-26T22:40:05

# cos x = 1/2 pm sqrt( 3)/2#

Is it: #1 - sin^2 x - cos x = 1/2# ??

If so, use the Pythagorean Identity #sin^2 t + cos^2 t = 1# to get:

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

Use algebra to turn that into a Quadratic:

# cos^2 x - cos x - 1/2 = 0#

This now looks like : #z^2 - z - 1/2 = 0#, right?!

So you can use the Quadratic Formula or just complete the square as follows:

# (z - 1/2 )^2 - 1/4 - 1/2 = 0#

# (z - 1/2 )^2 = 3/4#

# z = 1/2 pm sqrt( 3)/2#

# implies cos x = 1/2 pm sqrt( 3)/2#

Put that into your calculator. You won't get solutions for the plus bit of the #pm#.