# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 29

There are 3 values of $\theta$ in need to find, which is {$90^\circ, 210^\circ, 330^\circ$}.

#### Work Step by Step

$$2\sin\theta-1=\csc\theta$$ $$2\sin\theta-1=\frac{1}{\sin\theta}$$ $$2\sin\theta-\frac{1}{\sin\theta}-1=0$$ $$\frac{2\sin^2\theta-1-\sin\theta}{\sin\theta}=0$$ Now we divide both sides by $\sin\theta$, we have $$2\sin^2\theta-\sin\theta-1=0$$ Now you might use calculator or you can do something like this: $$2\sin^2\theta+\sin\theta-2\sin\theta-1=0$$ $$\sin\theta(2\sin\theta+1)-(2\sin\theta+1)=0$$ $$(2\sin\theta+1)(\sin\theta-1)=0$$ which means $$\sin\theta=-\frac{1}{2}$$ or $$\sin\theta=1$$ *For $\sin\theta=-\frac{1}{2}$ $\sin\theta$ is negative, which means the angles must lie in quadrant III and IV. Also, the reference angle would be $30^\circ$. 2 angles satisfy the above requirements which is {$210^\circ, 330^\circ$}. *For $\sin\theta=1$ $\sin\theta$ is positive, which means the angles must lie in quadrant I and II. Also, the reference angle would be $90^\circ$. 1 angle satisfy the above requirements which is {$90^\circ$}. Overall, there are 3 values of $\theta$ in need to find, which is {$90^\circ, 210^\circ, 330^\circ$}.

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