Question

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth, as appropriate.2sin θ ―1 = csc θ

Accepted Answer

Rewrite the given equation $$2\sin 	heta - 1 = \csc 	heta by $$expressing $$\csc 	heta in $$terms of $$\sin 	heta. $$Recall that $$\csc 	heta = \frac{1}{\sin 	heta}$$, so the equation becomes $$2\sin 	heta - 1 = \frac{1}{\sin 	heta}. $$Multiply both sides of the equation by $$\sin 	heta to $$eliminate the fraction, keeping in mind that $$\sin 	heta 
eq 0 to $$avoid division by zero. This gives $$2\sin^2$ 	heta - \sin 	heta = 1. $$Rearrange the equation to standard quadratic form in terms of $$\sin 	heta$$: $$2\sin^2$ 	heta - \sin 	heta - 1 = 0. $$Use the quadratic formula to solve for $$\sin 	heta. $$Recall the quadratic formula is $$x = \frac{-b \pm \sqrt{b^2$ - 4ac}}{2a}$$, where $$a=2$$, $$b=-1$$, and $$c=-1. $$Substitute these values to find the possible values of $$\sin 	heta. $$For each solution of $$\sin 	heta$$, determine the corresponding angles $$	heta in $$the interval $$[0^\circ, 360^\circ) by $$using the inverse sine function and considering the sine function's positive and negative values in different quadrants.

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth, as appropriate.
2sin θ ―1 = csc θ

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Key Concepts

Trigonometric Functions and Their Relationships

Solving Trigonometric Equations

Interval and Solution Constraints