## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 43

#### Answer

The solution set to this problem is $$\{\varnothing\}$$

#### Work Step by Step

$$\sin^2\theta-2\sin\theta+3=0$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$\sin^2\theta-2\sin\theta+3=0$$ Here we need to consider the equation a quadratic formula. In fact, if we take $\sin\theta$ as $x$, we would get $$x^2-2x+3=0$$ with $a=1, b=-2$ and $c=3$ - Calculate $\Delta$: $\Delta=b^2-4ac=(-2)^2-4\times1\times3=4-12=-8\lt0$ $\Delta\lt0$ means that there would be no values of $x\in R$ which can satisfy the given equation. That also means there would be no values of $\theta\in[0^\circ,360^\circ)$ which can satisfy the original equation $$\sin^2\theta-2\sin\theta+3=0$$ Therefore, the solution set to this problem is $$\{\varnothing\}$$

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