# trigonometric equation help!!!!!

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#1
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.
0
4 days ago
#2
(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.

0
#3
(Original post by Plücker)

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
0
4 days ago
#4
(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

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.
0
#5
(Original post by Plücker)
You can divide by 2 but this will not change the angle. The correct equation would then be

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 ??
0
4 days ago
#6
(Original post by prs33)
so it would be sin-1(-1/2)= -30 *2= -60 ??
It would be

What's the next step?
0
#7
(Original post by Plücker)
It would be

What's the next step?
ohhh , so θ= -15 ?
1
4 days ago
#8
(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.
1
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