Pre-cal

posted by .

1) Solve the equation

sinx(sinx+1) = 0

I am having trouble with the first step. Could I do sin x^2 + 1 =0?

  • Pre-cal -

    Using distributive property of multiplication over addition:
    p(p+1)=x²+x
    If p=sin(x), then
    sin(x)(sin(x)+1)=sin²(x)+sin(x)

    However, this particular problem does not require the expression to be expanded.
    In fact, if
    p(p+1)=0, it means that either p=0, or p+1=0.
    Substitute p=sin(x), we get:
    sin(x)=0 .......(1) or
    sin(x)+1=0.......(2)

    Solve for x in each of the two cases. The answer is the union of the two sets of answers.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Trigonometry.

    ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side …
  2. Trig........

    I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and …
  3. Trigonometic

    Solving Trigonometic Equations solve for x and give the answers as a equations : ( by radian) 1)cos(sinx)=1 <<<and thanks >>> We know sin 2x = 2(sinx)(cosx) so (sinx)(cos)=1/2(sin 2x) So we can change your equation …
  4. Math/Calculus

    Please take a look at my work below and provide a good critique: Solve the differential equation using the method of undetermined coefficients or variation of parameters. y'' - 3y' + 2y = sin(x) yc(x)= c1*e^2x+c2*e^x y"-3y'+2y=sin(x) …
  5. math

    the problem is 2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is what i've done so far: 2cos^2x+sinx-1=0 2cos^2x-1+sinx=0 cos2x + sinx =0 1 - 2sin^2x + sinx = 0 -2sin^2x+sinx-1=0 …
  6. calculus

    Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative
  7. Math - Pre- Clac

    Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1 - sinx + cosx)= (1 + sin x)/cosx Please and thankyou!
  8. Trig Help

    Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =?
  9. Math (trigonometric identities)

    I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx …
  10. Math Help

    Hello! Can someone please check and see if I did this right?

More Similar Questions