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sin 2x + cos x = 0 (substitute 2sin x cos x for sin 2x)
2sin x cos x + cos x = 0 (divide by cos x each term to both sides)
2sin x + 1 = 0 (subtract 1 to both sides)
2sin x = -1 (divide by 2 to both sides)
sin x = -1/2
Because the period of the sine function is 360⁰, first find all solutions in [0, 360⁰].
Because sin 30⁰ = 1/2 , the solutions of sin x = -1/2 in [0, 360] are
x = 180⁰ + 30⁰ = 210⁰ (the sine is negative in the third quadrant)
x = 360⁰ - 30⁰ = 330⁰ (the sine is negative in the fourth quadrant)
Thus, the solutions of the equation are given by
x = 210⁰ + 360⁰n and x = 330⁰ + 360⁰n, where n is any integer.
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If in a triangle tanA plus tanB plus tanC equals 0 then cotAcotBcotC equals?
cotA*cotB*cotC = 1/[tanA+tanB+tanC]
How do you construct the Lagrange interpolation for a polynomial cosx plus sinx given the four values of x?
Construct the Lagrange interpolating polynomial P1(x) for f(x) = cos(x)+sin(x) when x0 = 0; x1 = 0:3. Find the absolute error on the interval [x0; x1].
What is tan of 0?
Tan of 0 equals zero.
2 cos x plus 1 cos x plus -1 equals 0?
2 cos * cos * -1 = 2cos(square) * -1 =cos(square) + cos(square) *-1 =1- sin(square) +cos(square) * -1 1 - 1 * -1 =0
How do you solve the equation h equals -16t2 plus 64t plus 1224 for h?
-16t2 + 64t + 1224 = 0 Multiply both sides by -1 16t2 - 64t - 1224 = 0 Divide both side by 8 2t2 - 8t - 153 = 0 Cannot be factored so use the formula (-b (+ or -)(root of b2 - 4ac)) / 2a
Sin2x - radical 2 cosx equals 0?
Sin2x = radical 2
sinx+sin2x=0?
Sinx = 0 CosX= -1/2
What is the derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
Sinx plus cosx equals 0?
x = 3pi/4
How do you break 1 sinx divided 1-cosx?
0
How do you find the solutions of tanx equals 2cscx?
tanx=2cscx sinx/cosx=2/sinx sin2x/cosx=2 sin2x=2cosx 1-cos2x=2cosx 0=cos2x+2cosx-1 Quadratic formula: cosx=(-2±√(2^2+4))/2 cosx=(-2±√8)/2 cosx=(-2±2√2)/2 cosx=-1±√2 cosx=approximately -2.41 or approximately 0.41. Since the range of the cosine function is [-1,1], only approx. 0.41 works. So: cosx= approx. 0.41 Need calculator now (I went as far as I could without one!) x=approx 1.148
How do you solve sin squared theta plus cos theta equals sin theta plus cos squared theta?
For simplicity's sake, X represent theta. This is the original problem: sin2x+ cosX = cos2X + sinX This handy-dandy property is key for all you trig fanatics: sin2x+ cos2x = 1 With this basic property, you can figure out that sin2 x=1-cos2x and cos2x= 1-sin2x So we can change the original problem to: 1-cos2x+cosx = 1-sin2X + sinX -cos2x + cosx =-sin2x + sinX Basic logic tells you that one of two things are happening. sin2x is equal to sinx AND cos2x is equal to cosx. The only two numbers that are the same squared as they are to the first power are 1 and 0. X could equal 0, which has a cosine of 1 and a sine of 0, or it could equal pi/2, which has a cosine of 0 and a sine of 1. The other possibility whatever x (or theta) is, it's sine is equal to its cosine. This happens twice on the unit circle, once at pi/4 and once at 5pi/4. If you're solving for all possible values for x and not just a set range on the unit circle, then the final solution is: x=0+2pin x=pi/2+2pin x= pi/4 +2pin x=5pi/4+2pin (note that n is a variable)
What is the solution sets for sin3x plus sin2x plus sinx equals 0?
The solitions are in degrees. You may convert them to degrees should you wish. x= 0,90,120,180,240,270,360
Cos x plus sin x equals 0?
cosx + sinx = 0 when sinx = -cosx. By dividing both sides by cosx you get: sinx/cosx = -1 tanx = -1 The values where tanx = -1 are 3pi/4, 7pi/4, etc. Those are equivalent to 135 degrees, 315 degrees, etc.
How would you find x when 0 equals 2sinxcosx-cosx?
2sinxcosx-cosx=0 Factored : cosx(2sinx-1)=0 2 solutions: cosx=0 or sinx=.5 For cosx=0, x=90 or 270 degrees For sinx=.5, x=30 degrees x = {30, 90, 270}
Is the solution of 1 plus cosx equals 0 is π plus 2kπ?
Yes, that looks good. That's 180 degrees plus every multiple of 360 degrees more.
How do you solve sin2x plus 3 cos 2x equals 0?
sin2x + 3*cos2x = 0sin2x = -3*cos2xtan2x = -32x = arctan(-3)x = 0.5*arctan(-3) in the domain which should have been specified. As none has, the question has no answer.
