
Find the principal and general solutions of the following equations:
$(i)\;{\text{tanx = }}\sqrt 3 $
$(ii)\;{\text{secx = 2}}$
$(iii){\text{ cotx = - }}\sqrt 3 $
$(iv){\text{ cosecx = - 2}}$
Answer
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Hint: Here we will use the All STC rule to find the given tangent function’s quadrant and find the respective angle for the given value of the function. Then convert degrees into the form of “pi” for the principal and the general solution.
Complete step-by-step answer:
$(i)\;{\text{tanx = }}\sqrt 3 $
Given that - ${\text{tanx = }}\sqrt 3 $
By referring the trigonometric value table, we get ${\text{tan60}}^\circ {\text{ = }}\sqrt 3 $
Since, here given tangent function is positive.
And by All STC law, tangent is positive in first and the third quadrant.
The values for the tangent function in the first quadrant will be $ = 60^\circ $
The values for the tangent function in the third quadrant will be $ = 240^\circ + 60^\circ = 300^\circ $
Therefore the principal solutions are –
$x = 60^\circ {\text{ and }}x = 240^\circ {\text{ }}$
Convert the degree into radians.
$x = 60 \times \dfrac{\pi }{{180}}{\text{ and }}x = 240 \times \dfrac{\pi }{{180}}{\text{ }}$
Simplify the above equations and remove the common multiples from the numerator and the denominator.
$x = \dfrac{\pi }{3}{\text{ and }}x = \dfrac{{4\pi }}{3}{\text{ }}$
Hence, the principal solutions are - $x = \dfrac{\pi }{3}{\text{ and }}x = \dfrac{{4\pi }}{3}{\text{ }}$
General solution:
Let us Assume that $\tan x = \tan y$ and
Also $\tan x = \sqrt 3 $
From the above two equations, we get –
$\tan y = \sqrt 3 $
Since, we have calculated $\tan \dfrac{\pi }{3} = \sqrt 3 $while calculating the principal solutions.
$\tan y = \tan \dfrac{\pi }{3}$
The above equation –
$ \Rightarrow y = \dfrac{\pi }{3}$
Also, as per our assumption –
$\tan x = \tan y$
The general solution is –
$x = n\pi + y,\,{\text{where n}} \in {\text{z}}$
Place, $y = \dfrac{\pi }{3}$ in the above equation –
$ \Rightarrow x = n\pi + \dfrac{\pi }{3},\,{\text{where n}} \in {\text{z}}$
Hence, the general solution of $\tan x\,\;{\text{is }}x = n\pi + \dfrac{\pi }{3},\,{\text{where n}} \in {\text{z}}$
Note: Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ $ ).
Complete step-by-step answer:
$(i)\;{\text{tanx = }}\sqrt 3 $
Given that - ${\text{tanx = }}\sqrt 3 $
By referring the trigonometric value table, we get ${\text{tan60}}^\circ {\text{ = }}\sqrt 3 $
Since, here given tangent function is positive.
And by All STC law, tangent is positive in first and the third quadrant.
The values for the tangent function in the first quadrant will be $ = 60^\circ $
The values for the tangent function in the third quadrant will be $ = 240^\circ + 60^\circ = 300^\circ $
Therefore the principal solutions are –
$x = 60^\circ {\text{ and }}x = 240^\circ {\text{ }}$
Convert the degree into radians.
$x = 60 \times \dfrac{\pi }{{180}}{\text{ and }}x = 240 \times \dfrac{\pi }{{180}}{\text{ }}$
Simplify the above equations and remove the common multiples from the numerator and the denominator.
$x = \dfrac{\pi }{3}{\text{ and }}x = \dfrac{{4\pi }}{3}{\text{ }}$
Hence, the principal solutions are - $x = \dfrac{\pi }{3}{\text{ and }}x = \dfrac{{4\pi }}{3}{\text{ }}$
General solution:
Let us Assume that $\tan x = \tan y$ and
Also $\tan x = \sqrt 3 $
From the above two equations, we get –
$\tan y = \sqrt 3 $
Since, we have calculated $\tan \dfrac{\pi }{3} = \sqrt 3 $while calculating the principal solutions.
$\tan y = \tan \dfrac{\pi }{3}$
The above equation –
$ \Rightarrow y = \dfrac{\pi }{3}$
Also, as per our assumption –
$\tan x = \tan y$
The general solution is –
$x = n\pi + y,\,{\text{where n}} \in {\text{z}}$
Place, $y = \dfrac{\pi }{3}$ in the above equation –
$ \Rightarrow x = n\pi + \dfrac{\pi }{3},\,{\text{where n}} \in {\text{z}}$
Hence, the general solution of $\tan x\,\;{\text{is }}x = n\pi + \dfrac{\pi }{3},\,{\text{where n}} \in {\text{z}}$
Note: Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ $ ).
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