Answer :
⇒ 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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