SOLUTION: Solve for x on the interval [0,2pi].
2sin^2(x)-sin(x)=1
Someone please help, this assignment is due tomorrow and I cannot find how to do this anywhere.
Algebra ->
Trigonometry-basics
-> SOLUTION: Solve for x on the interval [0,2pi].
2sin^2(x)-sin(x)=1
Someone please help, this assignment is due tomorrow and I cannot find how to do this anywhere.
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Someone please help, this assignment is due tomorrow and I cannot find how to do this anywhere.
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= . --->
= ,
Factor the left side:
(2sin(x)+1)*(sin(x)-1) = 0.
Now the original equation deploys in two separate independent equations
1. 2sin(x)+1 = 0 ---> sin(x) = ---> x = and/or x = .
2. sin(x) - 1 = 0 ---> sin(x) = 1 ---> x = .
Answer. x = , , x = .
