Question

If 2sin2β – cos2β = 2, then β is​

Answer 1

To find: the value of $\beta$

Step-by-step explanation:

Now, $2\:sin2\beta-cos2\beta=2$

$\Rightarrow 2\times 2\:sin\beta\:cos\beta-(cos^{2}\beta-sin^{2}\beta)=2(sin^{2}\beta+cos^{2}\beta)$

• since $cos2\beta=cos^{2}\beta-sin^{2}\beta$
• $sin2\beta=2\:sin\beta\:cos\beta$
• $sin^{2}\beta+cos^{2}\beta=1$

$\Rightarrow 4\:sin\beta\:cos\beta-cos^{2}\beta+sin^{2}\beta=2\:sin^{2}\beta+2\:cos^{2}\beta$

$\Rightarrow sin^{2}\beta-4\:sin\beta\:cos\beta+3\:cos^{2}\beta=0$

$\Rightarrow sin^{2}\beta-sin\beta\:cos\beta-3\:sin\beta\:cos\beta+3\:cos^{2}\beta=0$

$\Rightarrow sin\beta(sin\beta-cos\beta)-3\:cos\beta(sin\beta-cos\beta)=0$

$\Rightarrow (sin\beta-cos\beta)(sin\beta-3\:cos\beta)=0$

Therefore, $sin\beta-cos\beta=0$

$\Rightarrow tan\beta=1=tan45^{\circ}$

$\Rightarrow \beta=45^{\circ}$

Or, $sin\beta-3\:cos\beta=0$

$\Rightarrow tan\beta=3$

$\Rightarrow \beta=tan^{-1}\dfrac{1}{3}$

Answer: $\beta$ is $45^{\circ}$ or $tan^{-1}\dfrac{1}{3}$

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Answer 2

If 2sin²β – cos²β = 2 then β is 90°

Given : 2sin²β – cos²β = 2

To find : The angle β

Solution :

Step 1 of 2 :

Write down the given equation

Here it is given that

2sin²β – cos²β = 2

Step 2 of 2 :

Find the value of β

2sin²β – cos²β = 2

⇒ 2sin²β – ( 1 - sin²β ) = 2

⇒ 2sin²β - 1 + sin²β = 2

⇒ 3sin²β - 1 = 2

⇒ 3sin²β = 3

⇒ sin²β = 1

⇒ sinβ = 1

⇒ β = 90°

Thus we have β = 90°

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