Given equation: `sin2x+cosx=0`, or `2sinxcosx+cosx=0`, or `cosx(2sinx+1)=0` `rArr cosx=0` or `2sinx+1=0` If `cosx=0` then `x=(2n+1)pi/2`, where `n in z`, and `2sinx+1=0, rArr sinx=-1/2=sin(pi+pi/6)` `rArr sinx=sin(pi + pi/6)` `rArr sinx=sin(7pi)/(6)` `rArr x=npi+(-1)^(n)(7pi)/(6)`, where `n in Z` Therefore, the general solution of given equation is `x=npi+(-1)^(n)(7pi)/(6)` or `(2n+1)pi/2, n in Z` Ans.
If sin 2x+cosx=0 , then which among the following is /are true ? I. cosx =0 , II. sinx=-(1)/(2) , III x=(2n+1)(pi)/(2),ninZ , IV. x=npi+(-1)^(n)(7pi)/(6),ninZ
Let f(x) = |(2cos^2x, sin2x, -sinx), (sin2x, 2 sin^2x, cosx), (sinx, -cosx,0)|, the value of int_0^(pi//2){f(x) + f'(x)} dx, is
Let f(x) = |(2cos^2x, sin2x, -sinx), (sin2x, 2 sin^2x, cosx), (sinx, -cosx,0)|, the value of int_0^(pi//2){f(x) + f'(x)} dx, is (i)pi/2 (ii)pi (iii)(3pi)/2 (iv)2pi
int_(0)^(pi/6) sin2x . cosx dx
Recall that sinx + cosx =u (say) and sin x cosx =v (say) are connected by (sinx +cosx)^(2) = sin^(2)x + cos^(2)x+2sin cosx rArr u^(2) = 1+2v rArr v=(u^(2)-1)/(2) It follows that any rational integral function of sinx + cosx , and sinx cosx i.e., R(sinx + cosx, sinx cosx) , or in our notation R(u,v) can be transformed to R(u, (u^(2)-1)/2) . Thus, to solve an equation of the form R(u,v)=0 , we form a polynomial equation in u and than look for solutions. The solution set of sinx + cosx -2sqrt(2) sin x cosx=0 is completely described by
intdx/(sin^2x+cosx)
Solve the equation sin2x-12(sinx-cosx)+12=0
Let Delta(x)=|(cos^(2)x,cosxsinx,-sinx),(cosxsinx,sin^(2)x,cosx),(sinx,-cosx,0)| then int_(0)^(pi//2){Delta(x)+Delta'(x)]dx equals
