What is the solution of [math]1+\sin(x)[/math],[math]1+\cos(x)[/math] and [math]1+\tan(x)?[/math]
1+sinx=0 => sinx=-1 => x= 3pi/2(or 270°) 1+cosx=0=> cosx=-1=> x= pi (or 180°)1+tanx=0 => tanx=-1 => x = 3pi/4(or 135°) Note: 1.These are not the only solutions , if you have idea about periods then we may use appropriate periods to complete the answers.2. You don't need to memorize these values if you just know the graphs.
What is the solution of lim x→0 2x-sinx/tanx+x?
Using Taylor series.Lim x → 0+ (2x-sin(x))/(tan(x)+x)=Lim x → 0+ ½–3/80*x^4+O(x^6)==Lim x → 0+ ½–3/80*0^4+O(0^6)=Lim x → 0+ ½–0+O(0)==Lim x → 0+ ½–0=½=Lim x → 0- (2x-sin(x))/(tan(x)+x).So Lim x → 0 (2x-sin(x)/(tan(x)+x)=½.Or Lim x → 0 (2x-sin(x))/(tan(x)+x)==Lim x → 0 ((2x-sin(x))/sin(x))/((tan(x)+x)/sin(x))==Lim x → 0 ((2x/sin(x)-sin(x)/sin(x))/((tan(x)/sin(x)+x/sin(x))==Lim x → 0 (2x/sin(x)-1)/((1/cos(x)+x/sin(x))==Lim x → 0 (2*1-1)/(1/1+1)=Lim x → 0 (2-1)/(1+1)=Lim x → 0 (1)/(2)=½.©Other function shifting a bracket:Lim x → 0+ (2x-sin(x))/(tan(x))+x==Lim x → 0+ (2x/sin(x)-sin(x)/sin(x))/(tan(x)/sin(x))+x==Lim x → 0+ (2x/x)-1)/(1/cos(x))+x=Lim x → +0 (2-1)/(1/1)+0=Lim x → +0 1/1==Lim x → 0+ 1=1. Likewise Lim x → +0 (2x-sin(x))/(tan(x))+x=1 thereforLim x → 0 (2x-sin(x))/(tan(x))+x exists and equals 1.Or using Taylor seriesLim x → 0 (2x-sin(x))/(tan(x))+x==Lim x → 0 1+x-1/6*x^2–31/360*x^4–43/15120*x^6+O(x^7)==Lim x → 0 1+0-1/6*0^2–31/360*0^4–43/15120*0^6+O(0^7)==Lim x → 0 1+0-0–0–0+0=Lim x → 0 1=1.
What is the solution of sin^2*3/2(x) + Cos^2*3/2(x)?
[math]\displaystyle \sin^2\left(\dfrac{3 \theta}{2}\right) + \cos^2\left(\dfrac{3 \theta}{2}\right)[/math]To make this more obvious, let us make the following substitution:[math]\displaystyle u = \dfrac{3 \theta}{2}[/math]Therefore, the equation becomes this:[math]\displaystyle \sin^2(u) + \cos^2(u)[/math]Does this equation remind you of anything? There is this thing called the Pythagorean trigonometric identity:[math]\displaystyle \sin^2(\theta) + \cos^2(\theta) = 1[/math]So you can see based on this substitution, this formula takes that form.Therefore, [math]\displaystyle \sin^2\left(\dfrac{3 \theta}{2}\right) + \cos^2\left(\dfrac{3 \theta}{2}\right) = 1[/math]
Solve the following trigonometric equation: 2sin(x)-sin(x)cos(x)=0?
there are going to be 4 solutions, since the period of sin 2x is 180. This means there will be 2 complete cycles in 360 degrees. let u = 2x - 10 sin u = 1/2 ==> u = 30 or 150 the general solution will be u = 30 + 360n or u = 150 + 360n, where n is an integer and 360 is the period of sin u now sub back in 2x - 10 for u: 2x - 10 = 30 + 360n 2x = 40 + 360n x = 20 + 180n or 2x - 10 = 150 + 360n 2x = 160 + 360n x = 80 + 180n for sin 2x, we need to let n = 0 and 1 first solutions: x = 20 or 80 second solutions: x = 200 or 260 all solutions in [0 , 360): x = 20, 80, 200, or 260 let's check 260: sin(2*260 - 10) = sin (520 - 10) = sin 510 subtracting 360, sin 510 = sin 150, which we already know is 1/2 checks...
Find all solutions for the equation: sec x - tan x = cos x
Used identity: Pythagorean trigonometric identity: cos²θ + sin²θ = 1 cos²θ = 1 - sin²θ ———————————————————— Get in terms of sine and cosine to simplify: sec(x) - tan(x) = cos(x) 1/cos(x) - sin(x)/cos(x) = cos(x) NOTE: You can only cancel out the cos(x) in the denominators here if cos(x)≠0, so we will have to check for this result later. cos(x)/cos(x) - cos(x)sin(x)/cos(x) = cos(x)cos(x) Multiply through by cosine to get rid of the denominators: 1 - sin(x) = cos²(x) Chance cos² into sine using the Pythagorean trigonometric identity: 1 - sin(x) = 1 - sin²(x) 1 - sin(x) = 1 - sin²(x) Set equal to zero: sin²(x) + 1 - 1 - sin(x) = 0 sin²(x) - sin(x) = 0 Factor: sin(x) [ 1 - sin(x) ] = 0 ---- Zero factor property: sin(x) = 0 or 1 - sin(x) = 0 sin(x) = 1 ---- In radians: sin(x) is zero at: x = π·n sin(x) is 1 at x = π/2 + 2·π·n (where n is any integer) ---- Re-NOTE: Remember that cos(x)≠0 business? Check for that result here. Cosine is zero at π/2 and 3π/2 and all other complete rotations of those positions. That means that the π/2 result we got is invalid, leaving only the other one. ———————————————— Answer: x = π·n where n is any integer ---- In degrees, this is: x = 180°·n where n is any integer
Does the following trigonometric equation have any solutions? If yes, obtain the solution [math]tan^{-1}\frac{x +1}{x-1} + tan^{-1}\frac{x-1}{x} = -tan^{-1} 7[/math] ?
We have,[math] \tan^{-1} \frac{x+1}{x-1}+\tan^{-1} \frac{x-1}{x}=-\tan^{-1} 7 \tag{i}[/math]Eq. (i) can be rewritten as,[math]\tan^{-1} \frac{x+1}{x-1} +\tan^{-1} \frac{x-1}{x}=\tan^{-1}(-7) [/math]Now, check out : Dhrubajyoti Bhattacharjee's answer to How do I solve tan-¹ (x+1) /(x-1) +tan-¹ (x-1) /x =tan-¹(-7)?[math] \boxed{x=2} [/math] -is not a root of the given equation.Hope, you'll understand..!!Thank You!
Solve sin^2 x - sin x - 2 = 0? 10 Points for Explanation?
The equation is a quadratic. It doesn't look like it at first. Use a substitute variable -- let's call it u u = sin(x) Then the equation sin^2(x) - sin(x) - 2 = 0 becomes: u^2 - u - 2 = 0 u = [ +1 +/- sqrt( 1 -4(1)(-2) ) ] / 2 u = [ 1 +/- sqrt( 1 + 8 ) ] / 2 u = [ 1 +/- sqrt(9) ] /2 u = [ 1 +/- 3 ] / 2 Using the + u = [ 1 + 3 ] / 2 = 4/2 However, since u = sin(x), this answer cannot be used in real numbers (sine cannot be bigger than 1). Using - u = [ 1 - 3 ] / 2 = -2/2 = -1 sin(x) = -1 x = - pi/2 (or -90 if you are working in degrees) --- Verification: x = -pi/2 sin(x) = -1 sin^2(x) = (-1)^2 = +1 sin^2(x) - sin(x) - 2 = 0 becomes +1 - (-1) - 2 = 0 1 + 1 - 2 = 0 Yep, it works --- The "lucky" approach: We know that sin(x) must always fall in the interval [-1, +1] therefore, sin^2(x) must belong to the interval [0, 1] We are given an equation where the sum two objects that can never be greater than 1, in absolute value, must equal + 2. sin^2(x) - sin(x) = 2 This forces us to give each one a limiting value the first one must be +1 (because a square is always positive), therefore the second one must be -1 The only logical solution MUST be sin(x) = -1 --- --"Can someone please explain where I went wrong" You took the second term as sin(x-2) You tried to solve: sin^2(x) - sin(x-2) = 0 which is not the same problem.
Find the values of sin x/2, cos x/2, and tan x/2 if 1)cos x= -1/3, x is in 3rd quadrant 2) tan x= 3/4, x is in 3rd quadrant?
cos(x) = -1/3 in the 3rd quadrant,Now use this formula cos(x) = 2cos^2(x/2) - 1,-1/3 = 2cos^2(x/2) - 1,cos^2(x/2) = 1/3 or cos(x/2) = 1/root(3), -1/root(3)Use sin^2(x/2) + cos^2(x/2) = 1,sin^2(x/2) = 1 - 1/3,sin(x/2) = root(2/3), -root(2/3)Since 180 < x < 270 i.e. x lies in the third quadrant, 90 < x/2 < 135,x/2 lies in the 2nd quadrant therefore cos and tan are negative, sin is positive,So cos(x/2) = -1/root(3), sin(x/2) = root(2/3), tan(x/2) = sin(x/2)/cos(x/2) = -22. tan(x) = 3/4 in 3rd quadrant,x lies in the 3rd quadrant and x/2 lies in the 2nd quadrant as seen in 1.tan(x) = 2tan(x/2)/[1-tan^2(x/2)],3/4 -3t^2/4 = 2t (assuming tan(x/2) = t),3t^2 +8t -3 = 0,Discriminant = 8^2 - 4(3)(-3) = 100,t = (-8 + 10)/2(3), (-8–10)/2(3) = 1/3, -3,Since x/2 lies in the 2nd quadrant, tan(x/2) = -3,tan(x/2) = sin(x/2)/cos(x/2) or sin(x/2) = -3cos(x/2) ….(i)Use sin^2(x/2) + cos^2(x/2) = 1 ….(ii)Substitute (i) in (ii),cos(x/2) = -1/root(10) and sin(x/2) = 3/root(10)
Solve this equation, 2cos^2 x - 1 = 0?
2 * cos² x - 1 = 0 <=> 2 * cos² x = 1 <=> cos² x = 1/2 <=> cos x = - sqr (1/2) or cos x = sqr (1/2) <=> cos x = - sqr (2) / 2 or cos x = sqr (2) / 2 <=> x = 3pi/4 or x = 5pi/4 or x = 7pi/4 or x = pi/4 Answers : pi/4 ; 3pi/4 ; 5pi/4 ; 7pi/4 ===> radians 45 ; 135 ; 225 ; 315 ===> degrees *******