Solve `cos 2x=|sin x|, x in (-pi/2, pi)`. – InterviewSolution

InterviewSolution

1.	Solve `cos 2x=\|sin x\|, x in (-pi/2, pi)`.
Answer» Correct Answer - `x=-pi/6, pi/6, (5pi)/6` If `sin x gt 0`, then we have `2 sin^(2)x+sin x -1 =0` `rArr sin x=1/2`, `:. x=pi/6, (5pi)/6` If `sin x lt 0` then we have `2sin^(2) x- sin x-1=0` `rArr sinx=-1/2` `:. x=-pi/6`

Discussion

No Comment Found

Related InterviewSolutions

If `xsina+ysin2a+zsin3a=sin4a``xsinb+ysin2b+zsin3b=sin4b``xsinc+ysin2c+zsin3c=sin4c`then the roots of the equation `t^3-(z/2)t^2-((y+2)/4)t+((z-x)/8)=0,a , b , c ,!=npi,`are`sina ,sinb ,sinc`(b) `cosa ,cosb ,cosc``sin2a ,sin2b ,sin2c`(d) `cos2a ,cos2bcos2c`A. `cos a, cos b, cos c`B. `sin a, sin b, sin c`C. `sin 2a, sin 2b, sin 2c`D. `cos 2a, cos 2b, cos 2c`
Consider the cubic equation `x^3-(1+cos theta+sin theta)x^2+(cos theta sin theta+cos theta+sin theta)x-sin theta. cos theta =0` Whose roots are `x_1, x_2 and x_3`A. 3B. 4C. 5D. 6
Cosider the cubic equation : `x^3-(1+costheta+sintheta)x^2+(costhetasintheta+costheta+sintheta)x-sinthetacostheta=0` whose roots are `x_1,x_2,x_3`. The value of `(x_1)^2+(x_2)^2+(x_3)^2` equalsA. 1B. 2C. `2 cos theta`D. `sin theta (sin theta+ cos theta)`
`tan^(3)x-3tanx=0`
If `tan(A-B)=1a n dsec(A+B)=2/(sqrt(3))`, then the smallest positive values of A and B, respectively, are`(25pi)/(24),(19pi)/(24)`(b) `(19pi)/(24),(25pi)/(24)``(31pi)/(24),(31pi)/(24)`(d) `(13pi)/(24),(31pi)/(24)`A. `(25 pi)/(24), (19 pi)/24`B. `(19 pi)/24, (25 pi)/24`C. `(31 pi)/24, (13 pi)/24`D. `(13pi)/24, (31 pi)/24`
Solve the equation:`cos^2[pi/4(sinx+sqrt(2)cos^2x)]-tan^2[x+pi/4tan^2x]=1`
Solve : `secx-tan x=sqrt(3)`.
`3^((1/2+log_3(cosx+sinx)))-2^(log_2(cosx-sinx))=sqrt(2)`
(i) `sinx=(sqrt(3))/(2)` (ii) `cosx=1` (iii) `secx=sqrt(2)`
(i) `sin2x=(1)/(2)` (ii) `cos3x=(1)/(sqrt(2))` (iii) `"tan"(2x)/(3)=sqrt(3)`