Question

If 2sin2β – cos2β = 2, then β is​

Answer 1

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$given \: 2 { \sin}^{2} \beta - { \cos }^{2} \beta = 2 \\ \\ - { \cos}^{2} \beta = 2 - 2 { \sin}^{2} \beta \\ \\ - { \cos }^{2} \beta = 2(1 - { \sin }^{2} \beta ) \\ \\ - { \cos }^{2} \beta = 2 { \cos }^{2} \beta \\ \\ 2 { \cos }^{2} \beta + { \cos }^{2} \beta = 0 \\ \\ 3 \ { \cos }^{2} \beta = 0 \\ \\ { \cos }^{2} \beta = \frac{0}{3} \\ \\ \cos\beta = \sqrt{0} \\ \\ so \: \cos \beta = 0$

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Answer 2

SOLUTION

TO DETERMINE

If 2sin²β – cos²β = 2 then β is

EVALUATION

Here it is given that

2sin²β – cos²β = 2

Now

2sin²β – cos²β = 2

⇒ 2sin²β – ( 1 - sin²β ) = 2

⇒ 2sin²β - 1 + sin²β = 2

⇒ 3sin²β - 1 = 2

⇒ 3sin²β = 3

⇒ sin²β = 1

⇒ sinβ = 1

⇒ β = 90°

FINAL ANSWER

If 2sin²β – cos²β = 2 then β is 90°

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