# If 3 sin x + 5 co

3 sin x +5cos x = 5

3sin x = 5-5cos x

3sin x = 5(1-cos x)

Squaring both sides we get

9sin2x = 25(1-cos x)2

9sin2x = 25(1+cos2x-2cos x)

9sin2x+9cos2x = 25 + 25cos2x - 50cos x + 9cos2x

9(sin2x+ cos2x) = 25 +34cos2x-50cos x

34cos2x-50cos x+16=0

17cos2x-25cos x+8=0

17cos2x-17cos x-8cos x+8=0

17cos x(cos x-1)-8(cos x-1)=0

When cos x = 1

3sin x + 5cos x = 5

3sin x = 0

Sin x = 0

Substituting the value cos x = 1 and sin x = 0

5(0)-3(1) = 0-3

-3.

sin2x+ cos2x = 1

sin2x = 1- cos2x

5sin x – 3cos x

-3 and 3.

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