Respuesta :
Answer:
Step-by-step explanation:
sin x =3/4
x is an obtuse angle and sin x is positive .
∴ x lies in 2nd quadrant.
Hence cos x is negative.
[tex]cos x=-\sqrt{1-sin^2x} =-\sqrt{1-(3/4)^2} =-\sqrt{1-\frac{9}{16} } =-\sqrt{\frac{7}{16} } =-\frac{\sqrt{7} }{4}[/tex]
Solving a trigonometric equation, it is found that:
[tex]\cos{x} = -\frac{\sqrt{7}}{4}[/tex]
The sine and the cosine of an angle x are related by the following equation:
[tex]\sin^2{x} + \cos^2{x} = 1[/tex]
In this problem:
- x is an obtuse angle, which means that it is the second quadrant, in which the sine is positive and the cosine is negative.
[tex]\sin{x} = \frac{3}{4}[/tex], thus:
[tex]\sin^2{x} + \cos^2{x} = 1[/tex]
[tex](\frac{3}{4})^2 + \cos^2{x} = 1[/tex]
[tex]\frac{9}{16} + \cos^2{x} = 1[/tex]
[tex]\cos^2{x} = 1 - \frac{9}{16}[/tex]
[tex]\cos^2{x} = \frac{7}{16}[/tex]
[tex]\cos{x} = \pm \sqrt{\frac{7}{16}}[/tex]
Since the cosine is negative.
[tex]\cos{x} = -\frac{\sqrt{7}}{4}[/tex]
A similar problem is given at https://brainly.com/question/24680641
