How do you solve 5sin2(x)+8sin⁡xcos⁡x−3=0? I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into sin⁡2x but then I have sin2x and sin⁡2x together.

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How do you solve 5sin2(x)+8sin⁡xcos⁡x−3=0?  I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into sin⁡2x but then I have sin2x and sin⁡2x together.

alexandrebaud43 · Accepted Answer

Hint:  5sin2xcos2x+8sin⁡xcos⁡x−3sin2x+3cos2xcos2x=0  2tan2x+8tan⁡x−3=0

Hattie Schaeffer · Accepted Answer

2(5sin2(x)+8sin⁡xcos⁡x−3)=5(1−cos⁡2x)+8sin⁡2x−6=−5cos⁡2x+8sin⁡2x−1=0  This is a classical linear trigonometric equation  8s=5c+1  64(1−c2)=(5c+1)2  c=−5±162289, s=5c+18

Vasquez · Accepted Answer

Rewrite it as 5sin2⁡x+8sin⁡xcos⁡x−3(sin2⁡x+cos2⁡x) ,which gives 2sin2⁡x+8sin⁡xcos⁡x−3cos2⁡x , a quadratic in sin⁡x Using the quadratic formula, the roots of 2u2+8u−3 are −2−112 and −2+112 Hence this is equivalent to (u+2+112)(u+2−112)=0 so the original factorises as (sin⁡x+(2+112)cos⁡x)(sin⁡x+(2−112)cos⁡x)=0

How do you solve 5\sin^2 (x)+8\sin x \cos x-3=0? I have

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