Question

Prove that the equation secx + cosecx + 2 = 0
has no real solution​

Answer 1

Step-by-step explanation:

Given:secx + cosecx + 2 = 0

To find: Solution of trigonometric equation

Solution:

We know that sec x= 1/cos x

and cosec x= 1/sinx

put these values to the expression

$\frac{1}{cos \: x} + \frac{1}{sin \: x} = - 2 \\ \\ \frac{sin \: x + \: cos \: x}{sinx \: cosx} = - 2 \\ \\ sin \: x + cos \: x = - 2sinx \: cosx \: \\\\\text{squaring both sides}\\ \\ {sin}^{2} x + {cos}^{2} x + 2sinxcosx = 4 {sin}^{2} x {cos}^{2} x \\ \\\text{apply identity}\:{sin}^{2} x+ {cos}^{2} x=1 \\\\ 1 + 2sinxcosx = 4 {sin}^{2} x {cos}^{2} x \\\\\text{apply identity} \:2sinx \: cosx=sin2x\\ \\1 + sin2x - ( {2sinx \: cosx})^{2} = 0 \\ \\ 1 + sin2x - {sin}^{2} 2x = 0 \\ \\ {sin}^{2} 2x - sin2x - 1 = 0$

It is a quadratic equation in sin2x

Put the value in quadratic formula to find the roots of equation

$sin2x_{1,2} = \frac{1 ± \sqrt{1 - 4(1)( - 1)}}{2} \\ \\ sin2x_{1,2} = \frac{1 ± \sqrt{5}}{2} \\ \\ sin2x_{1} = \frac{1 + \sqrt{5} }{2} \\ \\ 2x_1 = {sin}^{ - 1} \bigg(\frac{1 + \sqrt{5} }{2}\bigg) \\ \\ x_1 = \frac{1}{2} {sin}^{ - 1} \bigg(\frac{1 + \sqrt{5} }{2}\bigg) \\\\\text{it can't be value of x,}\\\text{because sin inverse is not defined for values greater than one} \\\\ x_2 = \frac{1}{2} {sin}^{ - 1} \bigg(\frac{1 - \sqrt{5} }{2}\bigg) \\ \\x_2=-19.02$

Hope it helps you.

To learn more on brainly:

1)if 21 cosec theta =29 find the value of cossquare theta-sinsquare theta/1-2sinsquare theta

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2)Prove this identity in trigonometry

https://brainly.in/question/21098065

Answer 2

Answer:

x²=-19.02 is your answer