Find the possible values of sin x, if 8 sin x-cos x = 4.

1.	Find the possible values of sin x, if 8 sin x-cos x = 4.
Answer» 8sin x - cos x = 4 Therefore, 8sin x - 4 = cos x We Know, sin^2 x + cos^2 x = 1…(Trigonometric Identity) Thus, sin^2 x + (8sin x - 4)^2 = 1 sin^2 x + 64 sin^2 x -64 sinx + 16 = 1 …(Since (a + b)^2 = a^2 + 2ab + b^2) Therefore, 65sin^2 x - 64 sin x + 16 = 1 65sin^2x - 64 sin x + 15 = 0 Solving this quadratic equation, we get, sin x = 3/5 and sin x = 5/13 Thus, possible values of sin x are: sin x = { 3/5 , 5/13}

Answer»

8sin x - cos x = 4

Therefore,

8sin x - 4 = cos x

We Know,

sin^2 x + cos^2 x = 1…(Trigonometric Identity)

Thus,

sin^2 x + (8sin x - 4)^2 = 1

sin^2 x + 64 sin^2 x -64 sinx + 16 = 1

…(Since (a + b)^2 = a^2 + 2ab + b^2)

Therefore,

65sin^2 x - 64 sin x + 16 = 1

65sin^2x - 64 sin x + 15 = 0

Solving this quadratic equation, we get,

sin x = 3/5 and sin x = 5/13

Thus, possible values of sin x are:

sin x = { 3/5 , 5/13}

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