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What is the general solution to the equation?

 

2sin^2x+5sinx−3=0

 Apr 9, 2020

Best Answer 

 #2
avatar+111329 
+1

2sin^2 x  + 5sin x  - 3    =  0      factor as

 

(2sin x   - 1) ( sin x   + 3)  =  0

 

The only  solution(s)  come(s) from setting the  first factor to 0 and solving

 

So   we  have

 

2sin x  - 1   = 0

 

2sin x  =  1

 

sin x  = 1/2

 

This happens  at   

 

pi/6  + n *2pi 

 

And

 

(5pi)/6  + n* 2 pi                where n is an integer

 

 

cool cool cool

 Apr 9, 2020
 #2
avatar+111329 
+1
Best Answer

2sin^2 x  + 5sin x  - 3    =  0      factor as

 

(2sin x   - 1) ( sin x   + 3)  =  0

 

The only  solution(s)  come(s) from setting the  first factor to 0 and solving

 

So   we  have

 

2sin x  - 1   = 0

 

2sin x  =  1

 

sin x  = 1/2

 

This happens  at   

 

pi/6  + n *2pi 

 

And

 

(5pi)/6  + n* 2 pi                where n is an integer

 

 

cool cool cool

CPhill Apr 9, 2020

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