SOLUTION: Solve the equation on the interval [0,360) {{{ 2Sin^2 (x)+ 3Sin(x)+1=0 }}} a)45(degrees) ,315(degrees) b)360(degrees) c)60(degrees),300(degrees) d)210(degrees),270(degrees),330

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the equation on the interval [0,360) {{{ 2Sin^2 (x)+ 3Sin(x)+1=0 }}} a)45(degrees) ,315(degrees) b)360(degrees) c)60(degrees),300(degrees) d)210(degrees),270(degrees),330      Log On


   

Question 928278: Solve the equation on the interval [0,360) +2Sin%5E2+%28x%29%2B+3Sin%28x%29%2B1=0+
a)45(degrees) ,315(degrees)
b)360(degrees)
c)60(degrees),300(degrees)
d)210(degrees),270(degrees),330(degrees)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
+2sin%5E2+%28x%29%2B+3sin%28x%29%2B1=0+


+2z%5E2%2B+3z%2B1=0+ Let z+=+sin%28x%29. That means z%5E2+=+sin%5E2%28x%29


Use the quadratic formula to solve +2z%5E2%2B+3z%2B1=0+. You should get these two solutions in terms of z


+z+=+-1%2F2 or z+=+-1


Now use each z solution to find the x solutions.


-------------------------------------------------------


If +z+=+-1%2F2, then,


z+=+sin%28x%29


-1%2F2+=+sin%28x%29


sin%28x%29+=+-1%2F2


x+=+arcsin%28-1%2F2%29 or x+=+180+-+arcsin%28-1%2F2%29


x+=+-30 or x+=+180+-+%28-30%29 Make sure you're in degree mode on your calculator.


x+=+-30 or x+=+210


We have x+=+-30 but the original interval for x is [0,360), so add 360 to -30 to get -30+360 = 330


So the two solutions here are x+=+330 or x+=+210


-------------------------------------------------------


If z+=+-1, then,


z+=+sin%28x%29


-1+=+sin%28x%29


sin%28x%29+=+-1


x+=+arcsin%28-1%29


x+=+270 Make sure you're in degree mode on your calculator.


There's only one solution here


-------------------------------------------------------


Put together, there are 3 solutions: x+=+210, x+=+270, x+=+330 and you're all done.


So the final answer is choice D
------------------------------------------------------------------------------------------------------------------------

If you need more one-on-one help, email me at jim_thompson5910@hotmail.com. You can ask me a few more questions for free, but afterwards, I would charge you ($2 a problem to have steps shown or $1 a problem for answer only).

Alternatively, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Any amount is greatly appreciated as it helps me a lot. This donation is to support free tutoring. Thank you.

Jim
------------------------------------------------------------------------------------------------------------------------