can someone solve this?
Given that
[tex] \sec(x) - 5 tan(x) = 3cos(x) [/tex]
show that
[tex]3 {sin(x) }^{2} - 5 sin(x) - 2 = 0[/tex]
Hence,find the solution,in radians,of the equation
[tex] \sec(x) - 5 \tan(x) =3 \cos(x) [/tex]
for
[tex] - \pi \leqslant x \geqslant \pi[/tex]

Respuesta :

9514 1404 393

Answer:

  x ≈ {-2.80176, -0.339837}

Step-by-step explanation:

Write in terms of sine and cosine:

  sec(x) -5tan(x) -3cos(x) = 0 . . . . . . given, subtract 3cos(x)

  1/cos(x) -5sin(x)/cos(x) -3cos(x) = 0

Multiply by cos(x). (Note, cos(x) ≠ 0.)

  1 -5sin(x) -3cos(x)² = 0

Use the trig identity to write in terms of sin(x).

  1 -5sin(x) -3(1 -sin²(x)) = 0

  3sin(x)² -5sin(x) -2 = 0 . . . . . . . . quadratic in sin(x)

  (sin(x) -2)(3sin(x) +1) = 0 . . . . . . factor the quadratic

Values of sin(x) that make this true are ...

  sin(x) = 2 . . . . . true only for complex values of x

  sin(x) = -1/3

Then the possible values of x are ...

  x = arcsin(-1/3), -π -arcsin(-1/3)

  x ≈ {-2.80176, -0.339837} . . . . . rounded to 6 sf

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