Question

2sin²A + 3cosA equal to zero

Answer 1

here we see that there are two trigonometric ratios, therefore in order to solve the equation we have to convert the trigonometric ratios into same form by replacing sin squareA by (1 - cos squareA)
2(1-cos^2A) + 3cosA = 0
or 2-2cos^2A + 3cosA = 0
or -2cos^2A + 3cosA + 2 = 0
or -2cos^2A +4cosA - cosA + 2 = 0
or -2cosA(cosA - 2) - 1(cosA - 2) = 0
or (-2cosA - 1)(cosA - 2) = 0
or -2cosA - 1 = 0
or -2cosA = 1
therefore cosA = 1/-2
0R
cosA - 2 = 0
therefore cosA = 2. inadmissible as maximum value of cos is 1.
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