Question

prove 4 sin^2x+6cos^2x=5​

Answer 1

Answer:

Given that

4cos^{2} x+6 sin^{2} x+5

This can be written as

⇒4cos ^{2}   x+4sin ^{2} x=5 2

⇒4(cos  ^{2 } x+sin^{2} x) + 2 sin^{2} x+5

⇒4+2sin  ^ {2}  x=5

⇒2sin  ^{2}  x=1

⇒sin  ^{2}  x= 1/2

⇒sinx= 1/sqrt{2}

⇒x=45 degrees

I hope this may helps you! :)

Answer 2

Answer:

x = 45°

Step-by-step explanation:

Correct Question :-

Solve 4sin²x + 6cos²x = 5

Given ,

4sin²x + 6cos²x = 5

To Find :-

Value of 'x'

Formula Required :-

1) sin²x + cos²x = 1

2) cos²x = (cosx)²

3)  cos45° = 1/√2

Solution :-

4sin²x + 6cos²x = 5

4sin²x + 4cos²x + 2cos²x = 5

[ ∴ 6cos²x can be written as '4cos²x + 2cos²x' ]

Taking '4' as common :-

4(sin²x + cos²x) + 2cos²x = 5

[ ∴ sin²x + cos²x = 1 ]

4(1) + 2cos²x = 5

4 + 2cos²x = 5

2cos²x = 5 - 4

2cos²x = 1

cos²x = 1/2

(cosx)² = 1/2

[ ∴ cos²x = (cosx)² ]

$cosx = \sqrt{\dfrac{1}{2}}$

$cosx=\dfrac{\sqrt{1}}{\sqrt{2}}$

cosx = 1/√2

cosx = cos45°

[ ∴ cos45° = 1/√2 ]

Cancelling 'cos' on both sides :-

x = 45°