Page 8 - 11-Math-14 Solutions of Trignometric Equation

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⇒ 1 - 2cos x + cos2 x =3sin2 x

⇒ 1 - 2cos x + cos2 x = 3(1 - cos2 x)

⇒ 4cos2 x - 2cos x - 2 =0

⇒ 2cos2 x - cos x -1 =0

⇒ (2cos x +1)(cos x -1) =0

⇒ cos x =- 1 or cos x =1
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i. If cos x = -1
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Since cos x is -v e in II and III Quadrants with the reference angle x=p
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⇒ x =p - p = 2p and x =p + p = 4p , where x U [0, 2p]
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Now x = 4p does not satisfy the given equation (i).
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∴ x =4p is not admissible and so x = 2p is the only solution.
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Since 2p is the period of cos x

∴ General value of x is 2p + 2np , nUZ
ii. If cos x = 1 3

⇒ x = 0 and x = 2p where x U [0, 2p]

Now both csc x and cot x are not defined for x = 0 and x = 2

∴ x = 0 and x = 2 are not admissible.

Hence solution set = 2p + 2np  , nUZ
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version: 1.1

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