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Question: How do you solve \( {\sec ^2}x = 4 \) in the interval \( 0 \leqslant x \leqslant 2\pi \)?...

How do you solve sec⁡2x=4{\sec ^2}x = 4 in the interval 0⩽x⩽2π0 \leqslant x \leqslant 2\pi?

Explanation

Solution

Hint : In the given question, we are required to find all the possible values of x that satisfy the given trigonometric equation sec⁡2x=4{\sec ^2}x = 4 lying in the range 0⩽x⩽2π0 \leqslant x \leqslant 2\pi . For solving such types of questions where we have to solve trigonometric equations, we need to have basic knowledge of algebraic rules and identities as well as a strong grip on trigonometric formulae and identities.

Complete step by step solution:
We have to solve the given trigonometric equation sec⁡2x=4{\sec ^2}x = 4 . We know that sec⁡(θ)=1cos⁡(θ)\sec \left( \theta \right) = \dfrac{1}{{\cos \left( \theta \right)}} .
So, converting sec⁡x\sec x into cos⁡x\cos x using the basic trigonometric formula, we get,
⇒1cos⁡2x=4\Rightarrow \dfrac{1}{{{{\cos }^2}x}} = 4
Now, Shifting the cos⁡x\cos x to right hand side of the equation and isolating the trigonometric ratio, we get,
⇒14=cos⁡2x\Rightarrow \dfrac{1}{4} = {\cos ^2}x
Taking square root both sides of the equation, we get,
⇒cos⁡x=±14\Rightarrow \cos x = \pm \sqrt {\dfrac{1}{4}}
⇒cos⁡x=±(12)\Rightarrow \cos x = \pm \left( {\dfrac{1}{2}} \right)
So, we have to find the values of x that satisfy either cos⁡x=(12)\cos x = \left( {\dfrac{1}{2}} \right) or cos⁡x=−(12)\cos x = - \left( {\dfrac{1}{2}} \right) lying in the range 0⩽x⩽2π0 \leqslant x \leqslant 2\pi.
We know that value of cos⁡(π3)=(12)\cos \left( {\dfrac{\pi }{3}} \right) = \left( {\dfrac{1}{2}} \right) and cos⁡(5π3)=(12)\cos \left( {\dfrac{{5\pi }}{3}} \right) = \left( {\dfrac{1}{2}} \right) .
Also, value of cos⁡(2π3)=−(12)\cos \left( {\dfrac{{2\pi }}{3}} \right) = - \left( {\dfrac{1}{2}} \right) and cos⁡(4π3)=−(12)\cos \left( {\dfrac{{4\pi }}{3}} \right) = - \left( {\dfrac{1}{2}} \right) .
Hence, the solutions for the trigonometric equation sec⁡2x=4{\sec ^2}x = 4 lying in the range 0⩽x⩽2π0 \leqslant x \leqslant 2\pi are: π3,2π3,4π3,5π3\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3} .

Note : Such trigonometric equations can be solved by various methods by applying suitable trigonometric identities and formulae. The general solution of a given trigonometric solution may differ in form, but actually represents the correct solutions. The different forms of general equations are interconvertible into each other.