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A student solved the equation sin2x/cos x = 2, 0 ≤ x ≤ pi, and got pi/2. What was the student's error?

Respuesta :

Answer:

Step-by-step explanation:

Albeit cosine is undefined at , this discontinuity is evitable by canceling terms. The expression is equivalent to , which is continuous for all . According to this result, the student did not make a mistake. Answer was correct.

Answer:

See below

Step-by-step explanation:

Since [tex]cos(\frac{\pi}{2})=0[/tex], the denominator will be 0, and the result would be indeterminate.

Do not read on if you do not understand Calculus

However, notice that [tex]\lim_{x \to \frac{\pi}{2}} \frac{sin(2x)}{cos(x)}=2[/tex] by L'Hopital's Rule:

[tex]\lim_{x \to \frac{\pi}{2}} \frac{sin(2x)}{cos(x)}\\\\\frac{2cos(2x)}{-sin(x)}\\ \\\frac{2cos(2(\frac{\pi}{2}))}{-sin(\frac{\pi}{2} )}\\\\\frac{2cos(\pi)}{-1}\\ \\\frac{-2}{-1}\\ \\2[/tex]