prs33
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I'm struggling on this question, someone pleaseee help me ;

2sin2θ+1=0

for 0∘≤θ≤360∘ are θ=

i got θ= 30, 120,210, 330 ; and got it wrong.
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Plücker
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(Original post by prs33)
I'm struggling on this question, someone pleaseee help me ;

2sin2θ+1=0

for 0∘≤θ≤360∘ are θ=

i got θ= 30, 120,210, 330 ; and got it wrong.
Judging from your answers you have made more than one mistake.

Post your working for feedback.
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prs33
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(Original post by Plücker)
Judging from your answers you have made more than one mistake.

Post your working for feedback.
From; 2sin2θ+1=0 i tried to solve it to sinθ+1/2=0 so sin-1(1/2)=30 so for my values I did 90+30=120, 180+30=210 , 360-30=330
i really need help
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Plücker
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(Original post by prs33)
From; 2sin2θ+1=0 i tried to solve it to sinθ+1/2=0 so sin-1(1/2)=30 so for my values I did 90+30=120, 180+30=210 , 360-30=330
i really need help
You can divide by 2 but this will not change the angle. The correct equation would then be \sin 2 \theta +\frac{1}{2}=0

Your next step is also incorrect. You need the inverse sine of both sides but only after you have isolated the sin term.

Subtract 1/2 from both sides then take the inverse sine.
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prs33
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(Original post by Plücker)
You can divide by 2 but this will not change the angle. The correct equation would then be \sin 2 \theta +\frac{1}{2}=0

Your next step is also incorrect. You need the inverse sine of both sides but only after you have isolated the sin term.

Subtract 1/2 from both sides then take the inverse sine.
so it would be sin-1(-1/2)= -30 *2= -60 ??
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Plücker
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(Original post by prs33)
so it would be sin-1(-1/2)= -30 *2= -60 ??
It would be 2\theta = \sin^{-1}\left(-\frac{1}{2}\right)

What's the next step?
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prs33
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(Original post by Plücker)
It would be 2\theta = \sin^{-1}\left(-\frac{1}{2}\right)

What's the next step?
ohhh , so θ= -15 ?
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Plücker
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(Original post by prs33)
ohhh , so θ= -15 ?
That is a solution of the equation. Now you just need to find all of the solutions in the interval 0<theta<360 degrees.
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