How to Solve Trigonometric Equation sin2x = sqrt(2)/2 Using Algebra?

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Discussion Overview

The discussion revolves around solving the trigonometric equation sin(2x) = sqrt(2)/2 using algebra, specifically within the interval from 0 to 2π. Participants explore how to derive all solutions for the equation.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents the equation sin(2x) = sqrt(2)/2 and mentions having found two solutions: π/8 and 3π/8, but seeks clarification on obtaining the additional solutions of 9π/8 and 11π/8.
  • Another participant notes the transformation of the interval for 2x, stating that if 0 ≤ x < 2π, then it follows that 0 ≤ 2x < 4π.
  • A participant confirms the calculations leading to the additional solutions, explaining that adding 2π to π/4 and 3π/4 yields 9π/4 and 11π/4, respectively, and dividing these by 2 results in 9π/8 and 11π/8.
  • Another participant reiterates the same calculations, affirming the correctness of the derived solutions.

Areas of Agreement / Disagreement

Participants appear to agree on the methods used to derive the additional solutions, with no evident disagreement on the calculations presented.

Contextual Notes

The discussion assumes familiarity with trigonometric identities and algebraic manipulation but does not explicitly address any potential limitations or assumptions underlying the solutions.

manish00333
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Solve sin2x= sqrt(2)/2 (using algebra)
the interval is between 0 and 2pi

i got the first two answers: pi/8 and 3pi/8, but i don t understand how to get the other two: 9pi/8 and 11pi/8.

thanks!
 
Last edited:
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If we are told:

$$0\le x<2\pi$$

then:

$$0\le 2x<4\pi$$
 
Oh, that makes sense!

So, pi/4 + 2pi = 9pi/4
and 3pi/4 + 2pi = 11pi/4

and when i divide them by 2, i get 9pi/8 and 11pi/8.

thank you so much! :D
 
manish00333 said:
So, pi/4 + 2pi = 9pi/4
and 3pi/4 + 2pi = 11pi/4

and when i divide them by 2, i get 9pi/8 and 11pi/8.

Perfect! (Yes)
 

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