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Question: How do you solve \(1+\cos x-2{{\sin }^{2}}x=0\) and find all solutions in the interval \(0\le x<360\...

How do you solve 1+cos⁡x−2sin⁡2x=01+\cos x-2{{\sin }^{2}}x=0 and find all solutions in the interval 0≤x<3600\le x<360 ?

Explanation

Solution

We are given a trigonometric equation in two trigonometric functions, sine and cosine function. In order to simplify this equation, we must apply basic trigonometric identities wahich have been derived and proven by application on the Pythagorean triangle. Then, we shall find solutions for the equation thus obtained in the specific given interval of x.

Complete step-by-step solution:
By the basic properties of trigonometric functions, we know that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1.
Transposing the term cos⁡2x{{\cos }^{2}}x to the right-hand side, we get the value of sin⁡2x{{\sin }^{2}}x as 1−cos⁡2x1-{{\cos }^{2}}x for substitution in the given equation, 1+cos⁡x−2sin⁡2x=01+\cos x-2{{\sin }^{2}}x=0.
We get, 1+cos⁡x−2(1−cos⁡2x)=01+\cos x-2\left( 1-{{\cos }^{2}}x \right)=0
Simplifying further, we get
⇒1+cos⁡x−2+2cos⁡2x=0 ⇒2cos⁡2x+cos⁡x−1=0 \begin{aligned} & \Rightarrow 1+\cos x-2+2{{\cos }^{2}}x=0 \\\ & \Rightarrow 2{{\cos }^{2}}x+\cos x-1=0 \\\ \end{aligned}
Let cos⁡x=t\cos x=t and putting in our equation, we get
⇒2t2+t−1=0\Rightarrow 2{{t}^{2}}+t-1=0
We will now factorize this equation to find the value of t.
⇒2t2+2t−t−1=0 ⇒2t(t+1)−1(t+1)=0 ⇒(t+1)(2t−1)=0 \begin{aligned} & \Rightarrow 2{{t}^{2}}+2t-t-1=0 \\\ & \Rightarrow 2t\left( t+1 \right)-1\left( t+1 \right)=0 \\\ & \Rightarrow \left( t+1 \right)\left( 2t-1 \right)=0 \\\ \end{aligned}
⇒t+1=0\Rightarrow t+1=0 or 2t−1=02t-1=0
⇒t=−1\Rightarrow t=-1 or t=12t=\dfrac{1}{2}
Thus, t=−1,12t=-1,\dfrac{1}{2}
Therefore, cos⁡x=−1,12\cos x=-1,\dfrac{1}{2}.
For these two values of x, we shall find the values of x which will satisfy our given range of x with the help of a graph of the cosine of x.

For cos⁡x=−1\cos x=-1,
In 0≤x<3600\le x<360, cos⁡x=−1\cos x=-1 at x=180x=180.
For cos⁡x=12\cos x=\dfrac{1}{2},
In 0≤x<3600\le x<360, cos⁡x=12\cos x=\dfrac{1}{2} at x=60,180+60x=60,180+60, that is, at x=60, 240x=60,\text{ }240.
Therefore, for 1+cos⁡x−2sin⁡2x=01+\cos x-2{{\sin }^{2}}x=0 all solutions in the interval 0≤x<3600\le x <360 are x=x= 60, 180 and 240.

Note: While simplifying any trigonometric equation, we must always try to simplify them as the sine or cosine functions. This is because only these two functions are the most basic of all the trigonometric functions as they have been derived straight from the Pythagorean theory. Also, we must remember the graphs of the six trigonometric functions to quickly find their solutions in between any interval as given in the question.