Solve the equation for exact solutions in the interval )[0°,360°)

use an algebraic method.
this is pretty simple but i can't figure it out.
i just need help working out the question.

the question is: 2sinθ-1=cscθ
the answers are 90°, 210°, 330°.

2sinθ-1= 1/sinθ
(2sinθ-1)*sinθ= (1/sinθ)*sinθ
2sin^2θ-sinθ= 1
sinθ(2sinθ-1)= 1

sinθ=1
θ=90°

2siθ°-1=1
2sinθ=2
sinθ=1
.... i can't get the answers 210° and 330°, which would make sinθ= -1/2

sorry about the extra ), that was a typo.

Quote:

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Originally Posted by somanyquestions

2sinθ-1= 1/sinθ
(2sinθ-1)*sinθ= (1/sinθ)*sinθ
2sin^2θ-sinθ= 1
sinθ(2sinθ-1)= 1

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from here,

$\displaystyle \sin \theta (2\sin\theta -1)= 1$

$\displaystyle 2\sin^2 \theta -\sin\theta = 1$

$\displaystyle 2\sin^2 \theta -\sin\theta - 1=0$

Now make $\displaystyle a =\sin \theta$ giving

$\displaystyle 2a^2 -a - 1=0$

Solve this quadratic.