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theta*asin(x) %2B asin(2*x)=pi/3 |
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Answer» sin^-1(x)+sin^-1(2x)=π/3 First, it is helpful if we rewrite this expression parametrically:a+b=π/3a=sin-1(x)b=sin-1(2x) In turn we can solve the second and third equations for x:x = sin(a)2x = sin(b) And then substitute sin(a) into 2x = sin(b) for:2sin(a) = sin(b) Now we have a system of two equations with two variable we can solve:a+b = π/32sin(a) = sin(b) Isolate a in the first equation:a = π/3-b And substitute:2sin(π/3-b) = sin(b) Use identity:sin(α-β)=sin(α)cos(β)-cos(α)sin(β) where we will use α=π/3 and β=b. This will change our expression to:2(sin(π/3)cos(b)-cos(π/3)sin(b) = sin(b)2((√3/2)cos(b)-(1/2)sin(b) =sin(b) √3cos(b)-sin(b)=sin(b) √3cos(b)=2sin(b) √3=2sin(b)/cos(b) √3/2=sin(b)/cos(b) tan(b)=√3/2 b=tan^-1(√3/2) Convert to inverseb≈.713724379 Since we already know that 2x=sin(b), we plug it in and get x≈.327326835. |
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