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Sagot :

The solutions to the equation [tex]\frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} = 3[/tex] are 20.8° and 69.1°

Note that:

[tex]\frac{sin \theta}{cos \theta} = tan \theta\\\\\frac{cos \theta}{sin \theta} =\frac{1}{tan \theta}[/tex]

The given equation is:

[tex]\frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} = 3[/tex]

Substitute [tex]\frac{sin \theta}{cos \theta} =tan \theta[/tex] and  [tex]\frac{cos \theta}{sin \theta} =\frac{1}{tan \theta}[/tex] into the equation [tex]\frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} = 3[/tex]

[tex]tan \theta + \frac{1}{tan \theta} = 3\\\\[/tex]

Multiply through by [tex]tan \theta[/tex]

[tex]tan^2\theta+1=3 tan \theta\\\\tan^2\theta-3 tan \theta + 1 = 0[/tex]

Let x = [tex]tan \theta[/tex]

[tex]x^2-3x+1=0\\\\[/tex]

x  =  2.62,  x  =  0.38

Since x = [tex]tan \theta[/tex]

[tex]2.62 = tan \theta\\\\\theta = tan^{-1}(2.62)\\\\\theta = 69.1^0[/tex]

[tex]0.38 = tan \theta\\\\\theta = tan^{-1}(0.38)\\\\\theta = 20.8^0[/tex]

Therefore, the solutions to the equation [tex]\frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} = 3[/tex] are 20.8° and 69.1°

Learn more on trigonometric equation here: https://brainly.com/question/14421002

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