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How do you solve #cosx=sinx#?

Answer 1

#pi/4 or 45^@#

Property of complementary arcs --> #cos x = sin (pi/2 - x)# There for, #sin x = sin (pi/2 - x)# a. #x = (pi/2 - x)# --> #2x = pi/2# --> #x = pi/4# b. #x = pi - (pi/2 - x) = pi/2 + x# (Rejected as non-sense)

Answer: #x = pi/4 or 45^@# Check. #x = pi/4# --> #sin x = sqrt2/2# --> #cos x = sqrt2/2# . OK

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Answer 2

Zoe Austin

To solve ( \cos(x) = \sin(x) ), you can use trigonometric identities to rewrite the equation in terms of one trigonometric function.

One approach is to use the Pythagorean identity: ( \sin^2(x) + \cos^2(x) = 1 ).

Rearranging this identity, we have: ( \sin^2(x) = 1 - \cos^2(x) ).

Substitute this expression for ( \sin^2(x) ) into the equation ( \cos(x) = \sin(x) ):

[ \cos(x) = \sqrt{1 - \cos^2(x)} ]

Now, solve for ( \cos(x) ):

[ \cos^2(x) = 1 - \cos^2(x) ]

[ 2\cos^2(x) = 1 ]

[ \cos^2(x) = \frac{1}{2} ]

Taking the square root of both sides:

[ \cos(x) = \pm \sqrt{\frac{1}{2}} ]

So, the solutions for ( x ) are ( x = \frac{\pi}{4} + 2n\pi ) and ( x = \frac{3\pi}{4} + 2n\pi ), where ( n ) is an integer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.