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# Trig question c2

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1. Solve for theta in the interval 0 < theta < 360
3sin^2theta+sintheta
Factorising this gives sintheta(3sintheta + 1)
sintheta = 0 or sintheta =-1/3

solutions for sin theta is the same as a sin graph , but how do i find the solutions of sintheta -1/3?
2. (Original post by Think People)
Solve for theta in the interval 0 < theta < 360
3sin^2theta+sintheta
Factorising this gives sintheta(3sintheta + 1)
sintheta = 0 or sintheta =-1/3

solutions for sin theta is the same as a sin graph , but how do i find the solutions of sintheta -1/3?
Do the inverse of sin (-1/3) to get a value for theta. Then use a CAST diagram or graphical methods to find the solutions.

Watch this if you don't know how to use a CAST diagram:
3. (Original post by Think People)
Solve for theta in the interval 0 < theta < 360
3sin^2theta+sintheta
Factorising this gives sintheta(3sintheta + 1)
sintheta = 0 or sintheta =-1/3

solutions for sin theta is the same as a sin graph , but how do i find the solutions of sintheta -1/3?
You do sine inverse to get the values of theta and draw a ASTC quadrant and see what values are valid for your range or you could just draw the sine graph and choose the valise which correspond to sin theta which are within your range
4. (Original post by hellomynameisr)
Do the inverse of sin (-1/3) to get a value for theta. Then use a CAST diagram or graphical methods to find the solutions.

Watch this if you don't know how to use a CAST diagram:
I tend to not use the CAST diagram method.
I draw the sin graph out then use symmetry to find the other points.For this type of question do you have any idea on how to do it without the cast method ?
5. (Original post by Shiv Loves Maths)
You do sine inverse to get the values of theta and draw a ASTC quadrant and see what values are valid for your range or you could just draw the sine graph and choose the valise which correspond to sin theta which are within your range
Inverse sin -1/3 gives -19.47... , where do i go from here
6. (Original post by Think People)
I tend to not use the CAST diagram method.
I draw the sin graph out then use symmetry to find the other points.For this type of question do you have any idea on how to do it without the cast method ?
so you draw the CAST diagram and then label the quadrants: q1 A, q2 S, q3 T, q4 C

since it is -1/3 and sin, you draw lines in q3 and 4

Then you can use the value of sin inverse (-1/3) to find the other solutions
7. (Original post by Think People)
I tend to not use the CAST diagram method.
I draw the sin graph out then use symmetry to find the other points.For this type of question do you have any idea on how to do it without the cast method ?
8. (Original post by Think People)
Inverse sin -1/3 gives -19.47... , where do i go from here
That fits within your range but there are other values which sin theta gives you 0 and - 1/3. It's sort of difficult to explain over tsr. You should watch h exam solutions videos on this subject
9. (Original post by Shiv Loves Maths)
That fits within your range but there are other values which sin theta gives you 0 and - 1/3. It's sort of difficult to explain over tsr. You should watch h exam solutions videos on this subject
I get it. The sin grpah i drew wasnt accurate.
-1/3 gives -0.3. I drew a line across and see which points meets with the graph.
Thx for the help

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