Math, asked by rsoniya605, 3 months ago

समीकरण 2sine=1 को हल कीजिए।​

Answers

Answered by s1282aditya114a
0

Answer:

sin−1x+sin−12x=sin−13–√2sin-1x+sin-12x=sin-132

⇒sin−1x−sin−1(3–√2)=−sin−1(2x)⇒sin-1x-sin-1(32)=-sin-1(2x)

⇒sin−1[x1−34−−−−−√−3–√41−x2−−−−−√]=sin−1(2x)⇒sin-1[x1-34-341-x2]=sin-1(2x)

⇒x2−3–√21−x2−−−−−√=−2x⇒x2-321-x2=-2x

⇒5x2=3–√21−x2−−−−−√⇒5x2=321-x2

⇒5x=3–√⋅1−x2−−−−−√⇒25x2=3(1−x2)⇒5x=3⋅1-x2⇒25x2=3(1-x2)

⇒28x2=3⇒x2=328⇒28x2=3⇒x2=328

∴x=±1237−−√∴x=±1237

परन्तु x=−1237−−√x=-1237 दी गयी समीकरण को संतुष्ट नहीं करता इसलिए x=1237−−√x=1237 दी गयी समीकरण का हल है ।

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