Find the exact values of the solutions in the interval 0 <= x < 2pi of the following equations without using a calculator. a) 3sec(x) + 2 = 8
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Answer:
a) [tex]x_{1} = \frac{\pi}{3}\,rad[/tex], [tex]x_{2} = \frac{5\pi}{3}\,rad[/tex], b) [tex]x_{1} = \frac{\pi}{4}\,rad[/tex], [tex]x_{2} = \frac{3\pi}{4}\,rad[/tex], [tex]x_{3} = \frac{5\pi}{4}\,rad[/tex], [tex]x_{4} = \frac{7\pi}{4} \,rad[/tex]
Step-by-step explanation:
a) We proceed to solve the expression by algebraic and trigonometrical means:
1) [tex]3\cdot \sec x + 2 = 8[/tex]
2) [tex]3\cdot \sec x = 6[/tex]
3) [tex]\sec x = 2[/tex]
4) [tex]\frac{1}{\cos x} = 2[/tex]
5) [tex]\cos x = \frac{1}{2}[/tex]
6) [tex]x = \cos^{-1} \frac{1}{2}[/tex]
Cosine has positive values in first and fourth quadrants. Then, we have the following two solutions:
[tex]x_{1} = \frac{\pi}{3}\,rad[/tex], [tex]x_{2} = \frac{5\pi}{3}\,rad[/tex]
b) We proceed to solve the expression by algebraic and trigonometrical means:
1) [tex]6\cdot \cos^{2} x = 3[/tex]
2) [tex]\cos^{2} x = \frac{1}{2}[/tex]
3) [tex]\cos x = \pm\frac{\sqrt{2}}{2}[/tex]
4) [tex]x = \cos^{-1} \left(\pm \frac{\sqrt {2}}{2} \right)[/tex]
There is one solution for each quadrant. That is to say:
[tex]x_{1} = \frac{\pi}{4}\,rad[/tex], [tex]x_{2} = \frac{3\pi}{4}\,rad[/tex], [tex]x_{3} = \frac{5\pi}{4}\,rad[/tex], [tex]x_{4} = \frac{7\pi}{4} \,rad[/tex]