Question

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$\sqrt{3} \sin(x)- \cos(x) = \sqrt{2}$
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Answer 1

||✪✪ QUESTION ✪✪||

solve it √3 sin x - cos x = √2 ?

|| ✰✰ ANSWER ✰✰ ||

→ √3 sin x - cos x = √2

Dividing both sides by 2 , we get,

→ (√3/2) sinx - (1/2) cosx = √2/2

Now, we know that :-

• (√3/2) = cos30°
• (1/2) = sin30°
• (√2/2) = (1/√2)

Putting These value now, we get,,

→ cos30° * sinx - sin30° cosx = 1/√2

→ sinx * cos30° - cosx * sin30° = (1/√2)

Now, comparing LHS with sin(A-B) = sinA*cosB - cosA*SinB , and putting (1/√2) = sin45° or sin135° in RHS, we get,

→ sin(x - 30°) = sin45° or sin135°

So, if ,

→ (x - 30°) = 45°

→ x = 75°

Or,

→ (x - 30°) = 135°

→ x = 165°

Hence, value of x can be 75° or 165°...