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sabika13
solve for x in the given interval sqrt2sinx+tanx=0 (-pi,pi)
even i'm waiting for someone to answer your question... eager to know ...
sabika only sin x is underroot or whole?
sqrsinx + tanx = 0 sqrsinx = - tanx sinx = tan^2x squaring both sides 1 = 1/cosx dividing both sides by sinx cosx = 1 crossmultiplying x = 0 i am not sure ... .. but i just tried i am also learning trigonometry
@sabika13 check my answer... i think i am so right.... tell me
sorry it was sqrt 2 sinx
1st time in my life i had tried such trig equ omg omg.... :( :(
sqrt 2sinx + tanx = 0 sqrt 2sinx = -tanx squaring both sides we get 2sinx = tan^2x 2 = sinx/cos^2x
sabika, before ppl can help you, they need to understand your question clearly. even now your question is vague... not clear..... is it sqrt2(sin x ) or sqrt(2sin x ) ?
@rizwan_uet can you keep going or are you stuck? @jatinbansalhot its sqrt2sinx
2cos*2x=sinx 2(1-sin^2x) = sinx 2-2sin^x -sinx = 0 sinx(-2sinx -1) = -2
-2sin^2x-sinx +2 = 0 is good! thanks!!
hhahahahaahahahahahahha... i'm not going to solve her problem... she is not clear...
? i dont know if its right.. thats why im asking a question..
up to this point i did my best but the rest is difficuilt for me sorry..
x = pi / 4, -pi / 4 .... m i right
sqrt 2sinx + tanx = 0 sqrt 2sinx = -tanx after squaring both sides we get 2sinx = tan^2x or sinx(2-sinx/(cos^2x))=0 then sinx=0 or x=0,pi,-pi (one part ) the other part may can be obtained from (2-sinx/(cos^2x))=0
how did you go from 2sinx=tan^2x to sinx(2-sinx/cos^2x0 =0 ?
yeah so sin^2x/cos^2x =2sinx 2sinxcos^2x=sinx^2 2sinx(1-sin^2x)=sin^2x ... am i doing it right?
2sinx cos^2x = sin^2x i have taken this statement from you 2sinx ( 1 - sin^2 ) = sin^2x 2sinx - 2sin^3 =sin^2x 2sinx - 2sin^3 - sin^2x = 0 2sin^3 + sin^2x - 2sinx = 0 sinx( 2sin^2 + sinx - 2 ) = 0
@sabika13 did my work help u? i want to know ..... do tell me