Question

If 0 < θ ≤ 90°, solve for 'θ' where

cos² θ - 3 cos θ + 2 = 2 sin² θ.​

Answer 1

Note : Theta is written as A.

Answer:

Required measure of angle A is 90°.

Step-by-step explanation:

= > cos^2 A - 3 cosA + 2 = 2 sin^2 A

From the properties of trigonometry :

• sin^2 ∅ + cos^2 ∅ = 1
• sin^2 ∅ = 1 - cos^2 ∅

= > cos^2 A - 3 cosA + 2 = 2( 1 - cos^2 A ) { 2sin^2 A = 2( 1 - cos^2 A ) }

= > cos^2 A - 3 cosA + 2 = 2 - 2 cos^2 A

= > 2 cos^2 A + cos^2 A - 3 cosA + 2 - 2 = 0

= > 3 cos^2 A - 3 cosA = 0

= > 3 cosA ( cosA - 1 ) = 0

Case 1 : If cosA is 0 .

= > 3 cosA = 0

= > cosA = 0

= > cosA = cos90°

= > A = 90° or π / 2

Case 2 : If cosA - 1 is zero.

= > cosA - 1 = 0

= > cosA = 1

= > cosA = cos0°

But here, theta or A is greater than 0°, cosA ≠ cos0°

Therefore the required measure of angle A is 90°.

Answer 2

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{90°}$

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Explanation—

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Identity —

sin²θ + cos²θ = 1

⇒sin²θ = 1 - cos²θ

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cos² θ - 3 cos θ + 2 = 2 sin² θ.

⇒cos² θ - 3 cos θ + 2 = 2(1-cos²θ)

⇒cos² θ - 3 cos θ + 2 = 2 - 2cos²θ.

⇒3cos²θ - 3cosθ = 0

⇒3cosθ(cosθ - 1) =0

3cosθ =0 ¦ cosθ - 1 =0

cos θ =0 ¦ cosθ =1

Now we know in 0 < θ ≤ 90°, θ is positive

∴ θ = 90 ¦ ∴ θ = 0

Satisfying this value in original equation

θ =90° satisfid✔

θ = 0° not satisfied ❌

Hence, the required angle is 90°

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