Sum of roots of equation x4−2x2sin2(πx2)+1=0 My try: x4−2x2sin2(πx2)+1=0∖x4+1=2x2(1−cos2(πx2)) (x2−1)2=−2x2cos2(πx2) (x2−1)2+2x2cos2(πx2)=0 x2−1=0,and,2x2cos2(πx2)=0

HINT:I'm sure you can solve the equation x2−1=0For the second equation, you have that eitherx2=0⇒x=0orcos⁡πx2=0⇒x=2π⋅(π2+πk)for some integer k.But can x=0?

"Sum of roots of equation x4−2x2sin2(πx2)+1=0 My try:..." solved and explained

amevaa0y

Answered question

2022-01-28

Sum of roots of equation

x^{4} - 2 x^{2} \sin^{2} (\frac{π x}{2}) + 1 = 0

My try:

x^{4} - 2 x^{2} \sin^{2} (\frac{π x}{2}) + 1 = 0 ∖ x^{4} + 1 = 2 x^{2} (1 - \cos^{2} (\frac{π x}{2}))

{(x^{2} - 1)}^{2} = - 2 x^{2} \cos^{2} (\frac{π x}{2})

{(x^{2} - 1)}^{2} + 2 x^{2} \cos^{2} (\frac{π x}{2}) = 0

x^{2} - 1 = 0, and, 2 x^{2} \cos^{2} (\frac{π x}{2}) = 0

Answer & Explanation

Karlie Wilkins

Beginner2022-01-29Added 7 answers

HINT:
I'm sure you can solve the equation

x^{2} - 1 = 0

For the second equation, you have that either

x^{2} = 0 \Rightarrow x = 0

\cos \frac{π x}{2} = 0 \Rightarrow x = \frac{2}{π} \cdot (\frac{π}{2} + π k)

for some integer k.
But can x=0?

Lily Thornton

Beginner2022-01-30Added 19 answers

Hint: This hint also handles complex roots. Let

a, b, c \in C

. Prove that, for a complex number

z, x = z

is a solution to

x^{4} - a, x^{2}, \sin^{2} (b x) + c = 0

if and only if x:=-z is a solution of the above equation.

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