How to solve this trig equation

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SUMMARY

The discussion focuses on solving the trigonometric equation 2sinθcosθ + 1 - 2sin²θ = 0. The user initially struggles with factoring and simplifying the equation but later recognizes the use of the double angle identity, sin(2x) = 2sin(x)cos(x). By rewriting the equation as -2sin²(x) + sin(2x) + 1 = 0, the user seeks guidance on further simplification and identity application for 1 - 2sin²(x).

PREREQUISITES
  • Understanding of trigonometric identities, specifically the double angle identities.
  • Familiarity with factoring quadratic equations.
  • Knowledge of the sine function and its properties.
  • Ability to manipulate algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Research the derivation and application of the double angle identity for sine.
  • Learn how to factor quadratic equations involving trigonometric functions.
  • Explore the identity for 1 - 2sin²(x) and its implications in trigonometric equations.
  • Practice solving various trigonometric equations to enhance problem-solving skills.
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Students studying trigonometry, educators teaching trigonometric identities, and anyone looking to improve their skills in solving trigonometric equations.

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Homework Statement



Solve for θ.
2sinθcosθ + 1 - 2sin^2θ = 0

Homework Equations





The Attempt at a Solution


I tried factoring, but I don't know how to continue with 2 diff ID's in one equation.
I don't know how to simplify it so that everything is in terms of one similar Identity.
:\
 
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Do you know any identities for 2sinθcosθ and 1 - 2sin2θ?
 
My mistake. You meant to show sin2(theta), and I misread this.
 
okay, I used 2sinxcos x and changed it to sin 2x so that I get:
-2sin^2(x) + sin2x + 1 = 0
How could I continue since I put everything in terms of sin?
 
Last edited:
Find an identity for 1 - 2sin2x and use that.
 

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