Solve 8sin⁡x=3cos⁡x+1sin⁡x My approach is as follow 8sin⁡x−1sin⁡x=3cos⁡x On squaring we get 64sin2x+1sin2x−16=3cos2x (64sin4x−16sin2x+1)(1−sin2x)=3sin2x Solving and re-arranging we get −64sin6x+80sin4x−20sin2x+1=0 Using the substitution sin2x=t −64t3+80t2−20t+1=0 I am not able to solve it from hence forth

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Solve 8sin⁡x=3cos⁡x+1sin⁡x  My approach is as follow  8sin⁡x−1sin⁡x=3cos⁡x  On squaring we get  64sin2x+1sin2x−16=3cos2x   (64sin4x−16sin2x+1)(1−sin2x)=3sin2x   Solving and re-arranging we get −64sin6x+80sin4x−20sin2x+1=0    Using the substitution sin2x=t   −64t3+80t2−20t+1=0    I am not able to solve it from hence forth

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First thing first, if you make the substitution t=sin2x the polynomial you get is−64t3+80t2−20t+1=0Now, to decompose it, you could use Ruffini&#039;s rule: first we find a zero of the polynomial that divides the constant term (in our case ±1). Let us call the polynomial P(t), thenP(1)=−64+80−20+1≠0P(−1)=64+80+20+1≠0clearly we have no integer solutions! So what we can do is substitute z=1t and what we get is the following polynomial in zQ(z)=z3−20z2+80z−64and now we apply the same rule: let us find a zero of Q(z) in the divisors of the constant term 64 which are ±1,±2,±4,⋯ You can easily see thatQ(4)=0then we can go on with the simplification and get(z−4)(z2−16z+16)=0which is definetly easier to solve. We findz1=4⇒t1=14z2=4(2−3)⇒t2=1{4(2−3)}z3=2(2+3)⇒t3=1{4(2+3)}from which you can find the values of sin2x

Solve 8 \sin x=\frac{\sqrt 3}{\cos x}+\frac{1}{\sin x} My approach is as

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