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May 15

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2cosx+2cos(2x)=0

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Consider the function
Locate and classify the function's stationary points.








doesn't work because tan is undefined at pi/2 --Alphador (talk) 10:17, 15 May 2010 (UTC)[reply]

Pi/2 is a potentially false solution since in the case of x=pi/2 dividing by cos x is dividing by zero. So this solution has to be tested. Setting x=Pi/2 into 2cos x+2cos 2x=2cos Pi/2+2cos pi=-2Taemyr (talk) 11:17, 15 May 2010 (UTC)[reply]
Convert into a quadratic in terms of . --COVIZAPIBETEFOKY (talk) 11:55, 15 May 2010 (UTC)[reply]
—Preceding unsigned comment added by 220.253.221.60 (talk) 12:14, 15 May 2010 (UTC)[reply]
The derivative of sin(2x) is not cos(2x). 76.229.218.70 (talk) 18:35, 15 May 2010 (UTC)[reply]
It is , as the OP has correctly written. -- Meni Rosenfeld (talk) 18:42, 15 May 2010 (UTC)[reply]
is another solution, and all solution are of course . In fact, all solutions can be written succinctly as . -- Meni Rosenfeld (talk) 18:42, 15 May 2010 (UTC)[reply]