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Question: Eliminate x, where \(p{{\sin }^{3}}x+q{{\cos }^{3}}x=\sin x.\cos x\) and \(p\sin x-q\cos x=0\) ....

Eliminate x, where psin3x+qcos3x=sinx.cosxp{{\sin }^{3}}x+q{{\cos }^{3}}x=\sin x.\cos x and psinxqcosx=0p\sin x-q\cos x=0 .

Explanation

Solution

  1. Substitute the value of psinxp\sin x from the second equation into the first equation and solve for q.
  2. Find the values of p and q in terms of sinx\sin x and/or cosx\cos x .
  3. Make use of the fact that sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 .

Complete step by step solution:
It is given that:
psin3x+qcos3x=sinx.cosxp{{\sin }^{3}}x+q{{\cos }^{3}}x=\sin x.\cos x ... (1)
psinxqcosx=0p\sin x-q\cos x=0 ... (2)
Using equation (2), we have:
psinx=qcosxp\sin x=q\cos x ... (3)
Substituting equation (3) in equation (1), we get:
qcosxsin2x+qcos3x=sinx.cosxq\cos x{{\sin }^{2}}x+q{{\cos }^{3}}x=\sin x.\cos x
qcosx(sin2x+cos2x)=sinx.cosxq\cos x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)=\sin x.\cos x
qcosx=sinx.cosxq\cos x=\sin x.\cos x
q=sinxq=\sin x ... (4)
Using equations (4) and (3), we get:
psinx=sinxcosxp\sin x=\sin x\cos x
p=cosxp=\cos x ... (5)
Finally, by squaring equations (4) and (5) and adding them together, we get:
p2+q2=sin2x+cos2x{{p}^{2}}+{{q}^{2}}={{\sin }^{2}}x+{{\cos }^{2}}x

p2+q2=1{{p}^{2}}+{{q}^{2}}=1 , which is the required answer.

Note:
Eliminating 'x' from two or more equations means that the equations are combined logically into a single equation so that it remains valid and 'x' does not appear in this new equation.
Elimination of a variable is used in converting parametric form to cartesian form.
If there are more unknowns than the number of equations, or more specifically, for an under-determined system of equations, one variable can be eliminated, usually by substitution.