How do you solve cos2x+sin2x=0

Question

This is a trig question solving for x using radians.

Answer 1

Question: solve cos2x+sin2x=0 .

cos2x+sin2x=0    --- rearrange

cos2x = -sin2x    --- divide both sides by cos2x

1 = -sin2x/tan2x, i.e.

tan2x = -1

Now tan(α) is positive if α is in the 1st quadrant or the 3rd quadrant.

But, since we have tan2x as being negative then 2x must lie in the 2nd quadrant or the 4th quadrant.

Since tan(-α) = -tan(α) and tan(π/4) = 1, then tan(-π/4) = -tan(π/4) = -1

i.e. 2x = -π/4, when 2x is in the 4th qaudrant

and 2x = 3π/4, when 2x is in the 3rd quadrant

Answer: x = -π/8, x = 3π/8

Answer 2

Given: cos2x+sin2x=0  … Eq.1.

     Let 2x be A (2x=A) and rewrite Eq.1 replacing 2x with A.

     We have: cosA+sinA=0.   Square both sides of the equation.

     We have: cos²A+2cosAsinA+sin²A=0.   Since cos²A+sin²A=1.

     We have: 1+2cosAsinA=0.  That is: 2cosAsinA=-1  …Eq.2.

     Simplify Eq.2 using one of the double-angle formulas.

     That is: 2cosAsinA=sin2A.   Plug this into Eq.2.

     We have: sin2A=-1.  …Eq.3.

     From Eq.3, We have: 2A=2n(pi)+3(pi)/2 (n: every integer).

     Since 2x=A, we have: 2x=2A/2={2n(pi)+3(pi)/2}/2.

     That is: 2x=n(pi)+3(pi)/4.  … Eq.4.    Plug Eq.4 into Eq.1.

     We have: cos{n(pi)+3(pi)/4}+sin{n(pi)+3(pi)/4}=0.…Eq.5

     CK: Eq.5 shows that Eq.1 occurs one time at every period of (pi).

     Here, we check Eq.5 for periods of (pi) caused by n=0 and 1.

     Case n=1 below is just a reassurance, because the period is (pi).

     1. If n=0, cos{3(pi)/4}+sin{3(pi)/4}=-1/(2^½)+1/(2^½)=0.

     2. if n=1, cos{(pi)+3(pi)/4}+sin{(pi)+3(pi)/4}

     =1/(2^½)-1/(2^½)=0.   The result satisfies Eq.5.   CKD.

      From Eq.4, we have: x=2x/2={n(pi)+3(pi)/4}/2.

      That is: x=n(pi)/2+3(pi)/8.

      Therefore, the answer satisfies Eq.1 is: x=n(pi)/2+3(pi)/8.