Solveeit Logo

Question

Question: Find the principal and general solutions of the following equations: \((i)\;{\text{tanx = }}\sqrt...

Find the principal and general solutions of the following equations:
(i)  tanx = 3(i)\;{\text{tanx = }}\sqrt 3
(ii)  secx = 2(ii)\;{\text{secx = 2}}
(iii) cotx = - 3(iii){\text{ cotx = - }}\sqrt 3
(iv) cosecx = - 2(iv){\text{ cosecx = - 2}}

Explanation

Solution

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)  tanx = 3(i)\;{\text{tanx = }}\sqrt 3
Given that - tanx = 3{\text{tanx = }}\sqrt 3
By referring the trigonometric value table, we get tan60 = 3{\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= 60^\circ
The values for the tangent function in the third quadrant will be =240+60=300= 240^\circ + 60^\circ = 300^\circ
Therefore the principal solutions are –
x=60 and x=240 x = 60^\circ {\text{ and }}x = 240^\circ {\text{ }}
Convert the degree into radians.
x=60×π180 and x=240×π180 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=π3 and x=4π3 x = \dfrac{\pi }{3}{\text{ and }}x = \dfrac{{4\pi }}{3}{\text{ }}
Hence, the principal solutions are - x=π3 and x=4π3 x = \dfrac{\pi }{3}{\text{ and }}x = \dfrac{{4\pi }}{3}{\text{ }}
General solution:
Let us Assume that tanx=tany\tan x = \tan y and
Also tanx=3\tan x = \sqrt 3
From the above two equations, we get –
tany=3\tan y = \sqrt 3
Since, we have calculated tanπ3=3\tan \dfrac{\pi }{3} = \sqrt 3 while calculating the principal solutions.
tany=tanπ3\tan y = \tan \dfrac{\pi }{3}
The above equation –
y=π3\Rightarrow y = \dfrac{\pi }{3}
Also, as per our assumption –
tanx=tany\tan x = \tan y
The general solution is –
x=nπ+y,where nzx = n\pi + y,\,{\text{where n}} \in {\text{z}}
Place, y=π3y = \dfrac{\pi }{3} in the above equation –
x=nπ+π3,where nz\Rightarrow x = n\pi + \dfrac{\pi }{3},\,{\text{where n}} \in {\text{z}}
Hence, the general solution of tanx  is x=nπ+π3,where nz\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  to 900^\circ \;{\text{to 90}}^\circ ) are positive, sine and cosec are positive in the second quadrant (90 to 18090^\circ {\text{ to 180}}^\circ ), tan and cot are positive in the third quadrant (180  to 270180^\circ \;{\text{to 270}}^\circ ) and sin and cosec are positive in the fourth quadrant (270 to 360270^\circ {\text{ to 360}}^\circ ).