Solve the given trigonometry equation

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SUMMARY

The discussion focuses on solving a trigonometric equation involving the tangent function and its periodicity. The user seeks clarification on why the cycles for sine and cosine are considered in addition to the standard π radians for tangent, which typically has a period of π. The solution involves the inverse tangent function, yielding values of -0.5071, 2.6345, and 5.7761, with the general solutions expressed as 2x = -0.5071 + nπ, 2x = 2.6345 + nπ, and 2x = 5.7761 + nπ, where n is an integer. The final solutions are constrained to the interval -10 ≤ x ≤ 0.

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chwala
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Homework Statement
see attached
Relevant Equations
Trigonometry
This is the problem. The question is simple i just need some clarification as indicated on the part highlighted below in red.

1662037532264.png
Now from my understanding tangent repeats on a cycle of ##π## radians...why do we have 2 the part circled in red below i.e ##2##? This is the part that i need clarity. Why consider cycles for sine and cosine here?
It's straightforward and less stressful to just use ##π##radians in my opinion.

1662037959983.png


The other form given below is the one that i know and am very much conversant with;

1662037901886.png
 
Last edited:
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We know that ##\tan^{-1}####\left[-\dfrac{5}{9}\right]=-0.5071##.

Tangent is negative in the second and fourth quadrant respectively. This will realize the given values: ##-0.5071+π## and ##2π-0.5071## respectively... giving us the desired ##2.6345## and ##5.7761##
i.e
##2x=-0.5071+nπ⇒x=-0.25355+\dfrac{1}{2}nπ##

##2x=2.6345+nπ⇒x=1.31725+\dfrac{1}{2}nπ##

##2x=5.7761+nπ⇒x=2.88805+\dfrac{1}{2}nπ##

where ##n=0, ±1, ±2...## and given that the solution lies on ##-10≤x≤0## then we shall get the answers as shown on the text.

Cheers.
 
Last edited:

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