How would you solve this equation? #cos^2Beta +cos Beta-2=0#

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2 Answers
May 2, 2018

#cos^2 beta + cos beta - 2 = 0#

#(cos beta -1)(cos beta +2 )= 0#

#cos beta = -2# isn't going to have any real solutions, and

#cos beta =1 # means #beta = 0^circ# in the range

#beta = 0^@#

Explanation:

Well, I think this method will be better.

Assume #cos beta = a#.

Now, The equation is,

#color(white)(xxx)a^2 + a - 2 = 0#

#rArr a^2 + (2 - 1)a - 2 = 0# [Break #1# as #2 - 1#]

#rArr a^2 + 2a - a -2 = 0# [Distributive Property]

#rArr a(a + 2) - 1(a + 2) = 0#

#rArr (a + 2)(a - 1)# [Grouping the like terms]

So, Either #a + 2 = 0 rArr a = -2#

Or, #a -1 = 0 rArr a = 1#

So, #cos beta = -2, 1#.

But cosine (#cos#) is a periodic function whose value ranges between #-1# and #1#. [#-1 <= cos beta <= 1#]

So, #cos beta != -2#.

And, #cos beta = 1#.

And We know, #cos (2npi) = 1#, where #n = ...-3, -2, -1, 0, 1, 2, 3....# so on.

So, for #n = 0#, #2npi = 0#,

That's why, #0^c# or #0^@# is a solution.

If #n = 1#, #2npi = 2pi#.

So, #2pi^c# or #360^@# is a solution too.

But, The interval #[0^@, 360^@)# is left-closed and right-open. So, #360^@# will not count.

So, #beta = 0^@#.

Hope this helps.