If #1/4# is a solution of #Sin2x=pCosx#, then p is? (a) -3 (b) 0 (c) 1 (d) -1 I think option (b) is correct, but I want solution.

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Answer 1 · 2016-12-28T14:34:14

#p=0.4948#

#sin2x=pcosx#

#hArr2sinxcosx=pcosx#

or #cosx(2sinx-p)=0#

i.e. either #cosx=0rArrx=(2n+1)pi/2#, where #n# is an integer - but this does not give us a value of #p#, as #1/4# is not among possible values.

or #2sinx-p=0# i.e. #p=2sinx#

as #1/4# is a solution, we have #2sin(1/4)-p=0# or #p=2sin(1/4)#

(assuming that #1/4# as solution is in radians)

#p=2xx0.2474=0.4948#