0
2 sin(x) - 3 = 0
2 sin(x) = 3
sin(x) = 1.5
No solution. The maximum value of the sine function is 1.0 .
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No, for example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero. It can't be proved because the statement is false. For example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero.
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I think you mean to solve:(-2 cos23x) - (3 sin 3x) = 0cos2 x + sin2 x = 1⇒ cos2x = 1 - sin2 x⇒ -2 cos2 3x - 3 sin 3x = -2(1 - sin2 3x) - 3 sin 3x = 0⇒ 2 sin2 3x - 3 sin 3x - 2 = 0⇒ (2 sin 3x + 1)(sin 3x - 2) = 0⇒ sin 3x = -1/2 or 2sin 3x = 2 is impossible as the range of sine is -1 ≤ sine ≤ 1Thus:sin 3x = -1/2⇒ 3x = 2nπ - π/6 or 2nπ - 5π/6⇒ x = 2/3nπ -π/18 or 2/3nπ -5π/18
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How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
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