How do you solve the following equation #tan(pi/4+(1/2))arccosx+tan(pi/4-(1/2))arccosx=1# in the interval [0, 2pi]?

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Answer 1 · 2016-07-07T02:39:15

The equation is in the form #arc cos x X oo=1# .

Use #tan( a+ b )=(tan a + tan b)/(1-tan a tan b)#

The given equation can be rearranged as

#arc cos x (tan (pi/4+1/2)+tan (pi/4-1/2))=1.#

The second factor takes the form #tan (pi/2)#/(non-zero constant).

So, the form is

arc cos x X #(+-oo)# = 1..

I have to stop here.

Note that is you use calculator ( in radian mode) for evaluating

#(tan (pi/4+1/2)+tan (pi/4-1/2))#, you end in syntax error.