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How do you solve #\sin x - 2\cos x - 1= 0#?

Answer 1

#90^@ + k360^@#
#216^@86 + k360^@#

sin x - 2cos x = 1 Call t that tan t = sin t/(cos t) = 2 Calculator gives #t = 63^@43#. Substitute 2 by #(sin t/cos t)# in the equation. #sin x - (sin t/cos t)cos x = 1# sin x.cos t - sin t.cos x = cos t = 0.447 Use trig identity: sin (x - t) = sin x.cos t - sin t.cos x. We get: #sin (x - 63.43) = 0.447# Calculator and unit circle give 2 solutions: a. #x - 63.43 = 26^@57# --> #x = 90^@ + k360^@# b. #x - 63.43 = 180 - 26.57 = 153^@43# --> #x = 153.43 + 63.43 = 216^@86 + k360^@#

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Answer 2

Jordan Armstrong

To solve the equation ( \sin(x) - 2\cos(x) - 1 = 0 ), you can use trigonometric identities to rewrite the equation in terms of a single trigonometric function. For example, you can express ( \sin(x) ) in terms of ( \cos(x) ) using the identity ( \sin^2(x) + \cos^2(x) = 1 ). Then, solve for ( \cos(x) ) and substitute it back into the original equation. Finally, solve for ( x ) using algebraic techniques.

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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.