Using the well-known trigonometric identity, we get: 1−coskx=2sin22kx. Thus, the function can be rewritten as: xtanx2sin22kx. Remind the well-known limit: xsinx→1,x→0. We rewrite the expression and receive: 4xk24k2x2tanx2sin22kx. Since, tanxx→1,x→0 and xsinx→1,x→0, we get: 4xk24k2x2tanx2sin22kx→4k2,x→0. From the latter and from the formulation of the task we get: 4k2=3. Finally, we get: k=±12. Thus, the answer is: k=±12.
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