Math and Arithmetic
Trigonometry

# What are the solutions of 2 cos squared x minus cos x equals 1?

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2cos2x - cosx -1 = 0

Factor:

(2cosx + 1)(cosx - 1) = 0

cosx = {-.5, 1}

x = {...0, 120, 240, 360,...} degrees

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## Related Questions Sin squared, cos squared...you removed the x in the equation. Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.  Writing x instead of theta, cos2x = 1 - (12/13)2 = 1 - 144/169 = 25/169 = (5/13)2 So cos(x) = &Acirc;&plusmn; 5/13 so that x = cos-1(5/13) or cos-1(-5/13) And then, depending on the range of x, you have solutions for x. A calculator will only give you the principal solutions, though. One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3 2 x cosine squared x -1 which also equals cos (2x) No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared) Since the word 'equals' appears in your questions it might be what is called a trigonometric identity, in other words a statement about a relationship between various trigonometric values. According to the Pythagorean identity, it is equivalent to sin2theta.   If x = sin &theta; and y = cos &theta; then: sin&sup2; &theta; + cos&sup2; &theta; = 1 &rarr; x&sup2; + y&sup2; = 1 &rarr; x&sup2; = 1 - y&sup2; Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals", "squared". It is not possible to be sure whether the second term is cos-squared of x or cos of 2x. The correct equation is a^3 - a^2*cos(a) - 6cos(a) = 0 which gives cos(a) = a^3/(a^2 + 6) This is not a simple equation to solve. In any case, before attempting a solution, it is necessary to know whether a is measured in degrees or radians. The information you provided in your question does not include an =. Therefore it is not an equation; it is an expression cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t) sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)  There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos &theta; + b sin &theta; = 8a sin &theta; - b cos &theta; = 5then square both sides of each to get:a&sup2; cos&sup2; &theta; + 2ab cos &theta; sin &theta; + b&sup2; sin&sup2; &theta; = 64a&sup2; sin&sup2; &theta; - 2ab sin &theta; cos &theta; + b&sup2; cos&sup2; &theta; = 25Now add the two together:a&sup2; cos&sup2; &theta; + a&sup2; sin&sup2; &theta; + b&sup2; sin&sup2; &theta; + b&sup2; cos&sup2; &theta; = 89&rarr; a&sup2;(cos&sup2; &theta; + sin&sup2; &theta;) + b&sup2; (sin&sup2; &theta; + cos&sup2; &theta;) = 89using cos&sup2; &theta; + sin&sup2; &theta; = 1&rarr; a&sup2; + b&sup2; = 89 Until an "equals" sign shows up somewhere in the expression, there's nothing to prove. Sin squared is equal to 1 - cos squared. (2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x No.Remember: sin2 + cos2 = 1So, in place of (1 - 2 sin2) we can write (sin2 + cos2 - 2 sin2).Massage that around slightly: (sin2 + cos2 - 2 sin2) = cos2 - sin2That's not equal to (sin2 - cos2), which is the original question we were asked. tan &theta; = sin &theta; / cos &theta; sec &theta; = 1 / cos &theta; sin &sup2; &theta; + cos&sup2; &theta; = 1 &rarr; sin&sup2; &theta; - 1 = - cos&sup2; &theta; &rarr; tan&sup2; &theta; - sec&sup2; &theta; = (sin &theta; / cos &theta;)&sup2; - (1 / cos &theta;)&sup2; = sin&sup2; &theta; / cos&sup2; &theta; - 1 / cos&sup2; &theta; = (sin&sup2; &theta; - 1) / cos&sup2; &theta; = - cos&sup2; &theta; / cos&sup2; &theta; = -1 sin2 + cos2 = 1 So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos) = (sin2 - cos2)/(sin*cos) = sin2/(sin*cos) - cos2/(sin*cos) = sin/cos - cos-sin = tan - cot

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