Solving a trigonometric equation

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Homework Help Overview

The discussion revolves around solving the trigonometric equation sin2x = -cos2x, with participants exploring various approaches and identities related to trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss initial steps taken to manipulate the equation, including attempts to factor expressions and the use of double angle identities. Questions arise about the correctness of these manipulations and the potential for alternative methods, such as solving a related equation siny = -cos y.

Discussion Status

The discussion is ongoing, with participants providing suggestions for alternative approaches and questioning the steps taken. There is an exploration of different identities that could simplify the problem, but no consensus has been reached on a specific method or solution.

Contextual Notes

Some participants express uncertainty regarding trigonometric identities and their applications, indicating a need for clarification on these concepts. There is also mention of specific values related to the equation, but the context remains exploratory.

angel_4_ever3
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I'm trying to solve:

sin2x = -cos2x

so my first step was...

2sinxcosx = -cos^2x - sin^2x

and then...

2sinxcosx + cos^2x + sin2x = 0

but I'm confused on how I would end up factoring 2sinxcosx + cos^2x + sin2x, or if I did something wrong earlier in the problem...thank you
 
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Can you find another trigonometric identity which would allow you to express the original equation in terms of either sine, or cosine, but not both? Maybe a different double angle identity?
 
angel_4_ever3 said:
I'm trying to solve:

sin2x = -cos2x

so my first step was...

2sinxcosx = -cos^2x - sin^2x

and then...

2sinxcosx + cos^2x + sin2x = 0

but I'm confused on how I would end up factoring 2sinxcosx + cos^2x + sin2x, or if I did something wrong earlier in the problem...thank you
Why not just solve siny= -cos y first and then find x?
 
HallsofIvy said:
Why not just solve siny= -cos y first and then find x?

I'm bad with identities, but wouldn't that be 3*pi/4?

square root of two over two equals negative negative of the same.
 

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