Respuesta :
The solution to the equations in the interval of (0, 2•π) are;
a. [tex]x = \frac{11 \cdot \pi }{6} [/tex]
b. [tex]x = \frac{\pi }{3} [/tex]
c. [tex]x = \frac{7 \cdot \pi }{6} [/tex][tex]x = \frac{11 \cdot \pi }{6} [/tex]
d. [tex]x = \frac{2 \cdot \pi }{3} [/tex]
[tex]x = \frac{ 5\cdot \pi }{6} [/tex]
How can trigonometric functions be evaluated?
The trigonometric functions can be expressed as follows;
a. 5•sin(x) + 1 = 3•sin(x)
5•sin(x) - 3•sin(x) = -1
2•sin(x) = -1
sin(x) = -1/2
The values of x are therefore;
[tex]x = - \frac{12 \cdot \pi \cdot n1 -7 \cdot \pi }{6} [/tex]
[tex]x = \frac{12 \cdot \pi \cdot n1 - \pi }{6} [/tex]
Which gives;
[tex]x = \frac{11 \cdot \pi }{6} [/tex]
b. 2•cos²(x) + cos(x) - 1 = 0
Factoring gives;
(2•cos(x) - 1)•(cos(x) + 1) = 0
2•cos(x) - 1 = 0
cos(x) = 1/2
Therefore;
[tex]x = - \frac{6 \cdot \pi \cdot n1 + \pi }{3} [/tex]
Which gives;
[tex]x = \frac{\pi }{3} [/tex]
c. -2•cos (2•x) = 2•sin (x)
-cos (2•x) = sin(x)
cos (2•x) = 1 - 2•sin²(x)
-(1 - 2•sin²(x)) - sin(x) = 0
2•sin²(x) - sin(x) - 1 = 0
(2•sin(x) + 1)•(sin(x) - 1) = 0
sin(x) = -1/2
[tex]x = - \frac{12 \cdot \pi \cdot n1 -7 \cdot \pi }{6} [/tex]
Which gives;
[tex]x = \frac{7 \cdot \pi }{6} [/tex]
Second part;
[tex]x = \frac{12 \cdot \pi \cdot n1 - \pi }{6} [/tex]
Which gives;
[tex]x = \frac{11 \cdot \pi }{6} [/tex]
d. csc(2•x) = -2/√3
sin(2•x) = -√3/2
Therefore;
[tex]x = - \frac{3 \cdot \pi \cdot n1 -2 \cdot \pi }{3} [/tex]
[tex]x = \frac{6 \cdot \pi \cdot n1 - \pi }{6} [/tex]
The values of x are;
[tex]x = \frac{2 \cdot \pi }{3} [/tex]
[tex]x = \frac{ 5\cdot \pi }{6} [/tex]
Learn more about trigonometric ratios here:
https://brainly.com/question/14421002
#SPJ1
