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Question: How do you solve \( {\sin ^2}x + 2\cos x = 2 \) over the interval \( 0 < x < 2\pi \) ?...

How do you solve sin⁡2x+2cos⁡x=2{\sin ^2}x + 2\cos x = 2 over the interval 0<x<2π0 < x < 2\pi ?

Explanation

Solution

Hint : In order to determine the solution of the above trigonometric equation in the interval 0<x<2π0 < x < 2\pi replace the sin⁡2x{\sin ^2}x using the identity of trigonometry sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x . Apply the formula (A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB by considering A as cos⁡x\cos x and B as 1. Derive the angle whose value of cosine is equal to 1 in the interval 0<x<2π0 < x < 2\pi to obtain the required answer.

Complete step by step solution:
We are given a trigonometric equation sin⁡2x+2cos⁡x=2{\sin ^2}x + 2\cos x = 2 and we have to find the solution between 0<x<2π0 < x < 2\pi
sin⁡2x+2cos⁡x=2{\sin ^2}x + 2\cos x = 2
Using the trigonometry sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x to replace sin⁡2x{\sin ^2}x from the equation ,we get
1−cos⁡2x+2cos⁡x=21 - {\cos ^2}x + 2\cos x = 2
Simplifying it further we have
⇒1−cos⁡2x+2cos⁡x−2=0 ⇒−cos⁡2x+2cos⁡x−1=0 ⇒cos⁡2x−2cos⁡x+1=0   \Rightarrow 1 - {\cos ^2}x + 2\cos x - 2 = 0 \\\ \Rightarrow - {\cos ^2}x + 2\cos x - 1 = 0 \\\ \Rightarrow {\cos ^2}x - 2\cos x + 1 = 0 \;
Now applying the formula of square of difference of two numbers (A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB by considering A as cos⁡x\cos x and B as 1
⇒(cos⁡x−1)2=0 ⇒cos⁡x−1=0 ⇒cos⁡x=1   \Rightarrow {\left( {\cos x - 1} \right)^2} = 0 \\\ \Rightarrow \cos x - 1 = 0 \\\ \Rightarrow \cos x = 1 \;
Taking inverse of cosine on both sides of the equation we get
cos⁡−1(cos⁡x)=cos⁡−1(1){\cos ^{ - 1}}\left( {\cos x} \right) = {\cos ^{ - 1}}\left( 1 \right)
Since cos⁡−1(cos⁡)=1{\cos ^{ - 1}}\left( {\cos } \right) = 1 as they both are inverse of each other
x=cos⁡−1(1)x = {\cos ^{ - 1}}\left( 1 \right)
The value of x is an angle between 0<x<2π0 < x < 2\pi whose cosine value is equal to 1.
As we know cos⁡0∘=1→0=cos⁡−1(1)\cos {0^ \circ } = 1 \to 0 = {\cos ^{ - 1}}\left( 1 \right)
x=0x = 0
Therefore, the solution to the given trigonometric equation is x=0x = 0 between 0<x<2π0 < x < 2\pi
So, the correct answer is “ x=0x = 0”.

Note : 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of cosine function is 2π2\pi .
3. The domain of cosine function is in the interval [0,π]\left[ {0,\pi } \right] and the range is in the interval [−1,1]\left[ { - 1,1} \right] .