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

sin(2x)/cos x = 2sin x*cos x/ cos x = 2 sin x = 2

sin x = 1

x = pi/2

but cos pi/2 = 0

so the given equation is indeterminate for x = pi/2

Answer with Step-by-step explanation:

We are given a equation:

Now, we will solve it

 2sinx=2

 sinx=1

all the possible values of x from 0 to π so that sinx=1 is

  x=π/2

but, cos(π/2) is indeterminate

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.