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General solution of :-
√3sinx+cosx=1
sinx+cosx=1
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उत्तर: -
√3 कोस x + पाप x = 1
√3sinx+cosx=1
समीकरण को 2 से विभाजित करें(Now we need to Divide the equation by 2=3/2)
- 2=3 / 2 कोस x + 1/2 पाप x = 1/2
- पाप π / 3 Cos x - Cos Sin / 3 पाप x = पाप π / 6
- पाप (> / 3 - x) = पाप 6/6
- पाप (x - π / 3) = पाप (-6/6) या पाप (π + π / 6)
- x -> / 3 = 2 n π - π / 6 या 2 n - + 7 6/6
x = 2 n = + π / 6 या 2 n 3 + 3 π / 2
यह 2 n 2 + 3 2/2 के रूप में भी व्यक्त किया जा सकता है: 2 n π + 2π + π / 2
यानी।, 2 मीटर π +, / 2
(This equation can be also expressed as this)
2 n 2 + 3 2/2: 2 n π + 2π + π / 2
2 meters π +, / 2
इसलिए, (Therefore, or Hence)
x = 2 n π + π / 6 और x = 2 n n - n / 2
दूसरा रास्ता:
- √3 कोस x +पाप x = 1
- =3 / 2 Cos x - 1/2 पाप x = 1/2
- Cos π / 6 Cos x + Sin 6/6 Sin x = Cos π / 3
- Cos (x + π / 6) = Cos π / 3
हमारे पास Cos X = Cos Y, X = 2 n or + Y या 2 n- Y के लिए सामान्य समाधान है
Cos X = Cos Y, X = 2 n or + Y or 2 n-y is. the common solution or general solution:
- x +> / 6 = 2 n π + π / 3 या 2 n + - π / 3
- x = 2 n π + π / 6 या 2 n π + n / 2
Hope my answer will be helpful to you and everyone :
TANU81:
Wonderful answer ⭐ but in Hindi I am unable to understand :(
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