The problem involves solving for θ in the trigonometric equation 2sinθcosθ + 1 - 2sin²θ = 0. Participants are exploring various methods to approach this equation, including the use of trigonometric identities and transformations.
The discussion is active with multiple approaches being considered. Some participants have provided guidance on using half and double angle identities, while others express confusion about specific steps in the reasoning. There is no explicit consensus on a single method, but various paths are being explored.
Some participants mention constraints such as not having learned certain identities, which may affect their ability to apply specific methods discussed in the thread.
Bohrok said:You also could write it as [itex]\cos 2x + \sin 2x = 0 \Longrightarrow sin2x = -cos2x \Longrightarrow tan2x = -1[/itex]
Vee9 said:The answer is 67.5.
So from the tan2x = -1 that you found, I did:
45+ (45/2) = 67.5
One thing I didn't understand was how you got
tan2x = -1 from sin2x = -cos2x ?
Thanks. Major mind block today, lol.
jambaugh said:There are multiple paths.
Apply both half/double angle identities, i.e. also [tex]\sin^2(\theta)=(1-\cos(2\theta))/2[/tex]
This gives you an equation you can solve directly because of its simple form.
One other approach is to not use half angle identities but to use the base pythagorean identity on the cosine. You'll get a square root so solve for that term and square it out. The result will be a quartic equation but its really a quadratic equation in sine squared of theta so you can solve it easily.