How do I find #x#: #2 sin x cos x-1=cos2x#?

Question

I tried doing it:

#(2 sin x cos x)-1=2 cos^2x -1#
#(2 sin x cos x)-2 cos^2 x=0#

And I'm not sure if what I'm doing is correct?

Asher Friedman · Accepted Answer

undefined To solve the equation $2 \sin(x) \cos(x) - 1 = \cos(2x)$ for $x$, you can use the double-angle identity for cosine, which states that:

$$ \cos(2x) = 2 \cos^2(x) - 1 $$

Substituting this into the equation gives:

$$ 2 \sin(x) \cos(x) - 1 = 2 \cos^2(x) - 1 $$

Now, we can simplify this equation:

$$ 2 \sin(x) \cos(x) = 2 \cos^2(x) $$

Divide both sides by 2:

$$ \sin(x) \cos(x) = \cos^2(x) $$

Now, we can use the identity $ \sin(2x) = 2 \sin(x) \cos(x) $:

$$ \sin(2x) = \cos^2(x) $$

Next, we use the identity $ \cos^2(x) = 1 - \sin^2(x) $:

$$ \sin(2x) = 1 - \sin^2(x) $$

Now, replace $ \sin(2x) $ with $ 2\sin(x)\cos(x) $:

$$ 2\sin(x)\cos(x) = 1 - \sin^2(x) $$

Now, rearrange the equation:

$$ \sin^2(x) + 2\sin(x)\cos(x) - 1 = 0 $$

This equation is a quadratic in terms of $ \sin(x) $. You can solve it using the quadratic formula:

$$ \sin(x) = \frac{-2\cos(x) \pm \sqrt{(2\cos(x))^2 - 4(-1)}}{2} $$

Simplify:

$$ \sin(x) = \frac{-2\cos(x) \pm \sqrt{4\cos^2(x) + 4}}{2} $$

$$ \sin(x) = \frac{-2\cos(x) \pm 2\sqrt{\cos^2(x) + 1}}{2} $$

$$ \sin(x) = -\cos(x) \pm \sqrt{\cos^2(x) + 1} $$

Now, you can solve for $x$ using the inverse sine function:

$$ x = \arcsin(-\cos(x) \pm \sqrt{\cos^2(x) + 1}) $$

Remember to check the domain of the arcsin function based on the values of $ \cos(x) $ to ensure you get all possible solutions for $ x $.

Carter Anderson · Answer

#x=pi/2+2pin, x=(3pi)/2+2pin, x=pi/4+pin#, where #n# is any integer

Assuming the equation you want to solve is this: #2sin(x)cos(x)-1=2cos^2(x)-1# #2sin(x)cos(x)=2cos^2(x)# #2sin(x)cos(x)-2cos^2(x)=0# Then we factor it: #2cos(x)(sin(x)-cos(x))=0# For the expression to be equal to zero, either the first, second or both factors need to be equal to zero, so let's look at when the factors are equal to #0#: #2cos(x)=0# #cos(x)=0/2=0# This equation actually has an infinite number of solutions since the #cos# function is periodic, so #x# can have these values, where #n# is any integer: #x=pi/2+2pin, x=(3pi)/2+2pin# Now, let's look at the zeroes for the other factor, #sin(x)-cos(x)#: #sin(x)-cos(x)=0# #(sin(x)-cos(x))/cos(x)=0/cos(x)# #sin(x)/cos(x)-1=0# #sin(x)/cos(x)=1# Using the identity that #sin(x)/cos(x)=tan(x)#, we get: #tan(x)=1# This function is once again periodic, so the above equation has an infinite number of solutions as well. They are these, where #n# is any integer: #x=pi/4+pin# Combining these with the ones from the other equation, we get that these are the possible solutions to the original equation (of course, assuming the trigonometric arguments are in radians): #x=pi/2+2pin, x=(3pi)/2+2pin, x=pi/4+pin#

Elijah Alexander · Answer

#x=pi/2+kpi;k inZZ,x=pi/4+-2pik;k inZZ# #"using the "color(blue)"trigonometric identity"# #•color(white)(x)cos2x=2cos^2x-1# #rArr2sinxcosx-1=2cos^2x-1# #rArr2sinxcosx-2cos^2x=0larrcolor(blue)"common factor of 2cosx"# #rArr2cosx(sinx-cosx)=0# #"equate each factor to zero and solve for x"# #•color(white)(x)cosx=0rArrx=pi/2# #"due to the periodicity of cos this value will repeat"# #"general solution is "x=pi/2+-kpitok inZZ# #•color(white)(x)sinx-cosx=0# #rArrsinx=cosxrArrx=pi/4# #"general solution is "x=pi/4+-2piktok inZZ#

How do I find #x#: #2 sin x cos x-1=cos2x#?

I tried doing it: #(2 sin x cos x)-1=2 cos^2x -1# #(2 sin x cos x)-2 cos^2 x=0# And I'm not sure if what I'm doing is correct?

I tried doing it:

#(2 sin x cos x)-1=2 cos^2x -1#
#(2 sin x cos x)-2 cos^2 x=0#

And I'm not sure if what I'm doing is correct?