Question

Hence solve 3 cos²2x/1+sin 2x = 1 for 0° < x < 90°.​

Answer 1

The value of x = 45°

It is given that 3 cos² 2x / ( 1 - sin 2x) = 1

3cos² 2x / ( 1 - sin 2x) = 1

As we know sin²x + cos²x = 1

3 ( 1 - sin²2x) / ( 1 - sin 2x) = 1

3 - 3 sin²2x = 1 - sin 2x

3 sin²2x - sin2x - 2= 0

As we know sin2x = 2 sin x. cos x

3( 2 sin x. cos x)² + 2 sin x. cos x  - 2 = 0

Let's take sin x . cos  x = t

3 ( 2t)² - 2t - 2 = 0

12t² - 2t - 2 = 0

12t² - 6t + 4t - 2 = 0

6t( 2t - 1) + 2(2t - 1) = 0

(6t + 2)(2t - 1) = 0

t = -1/3 , 1/2

Now put the value of t we get

sin x . cos x = -1/3 , 1/2

2 sin x . cos x = -2/3 , 1

sin 2x = -2/3 , 1

It is given that x should be between 0° to 90°.

Therefore

sin 2x = 1

sin 2x = sin 90°

2x = 90°

x = 45°

Therefore the value of x is 45°

To know more about the trigonometry value refer to the link given below:

brainly.com/question/25618616

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