Question

How do you solve \[2\cos x+1=0\] for \[{{0}^{\circ }}\le x\le {{360}^{\circ }}\]?

Answer 1

Hint: In this problem, we have to solve for x from the given trigonometric expression within the given interval. We can first subtract 1 on both sides and we can divide by 2 on both sides, we can get an expression which is equal to cos x. Then we can find the value for x for which the cos value equals to the given data.

Complete step by step solution:
We know that the given trigonometric expression is,
 \[2\cos x+1=0\]
 We can now subtract 1 on both the left-hand side and the right-hand side of the equation ,we get
\[\Rightarrow 2\cos x=-1\]
Now we can divide the above step by 2 on both the left-hand side and the right-hand side of the equation, we get
\[\Rightarrow \cos x=\dfrac{-1}{2}\]
We know that cos x is less than 0, then x is in the second and third quadrant.
We know that the value of x will be \[{{60}^{\circ }}\], when cos x = \[\dfrac{1}{2}\].
We can subtract 180 degree and 60 degree to get the value of x in second quadrant, we get
\[\Rightarrow x=\left( {{180}^{\circ }}-{{60}^{\circ }} \right)={{120}^{\circ }}\] in second quadrant.
We can add 180 degree and 60 degree to get the value of x in third quadrant, we get
\[\Rightarrow x=\left( {{180}^{\circ }}+{{60}^{\circ }} \right)={{240}^{\circ }}\] in third quadrant.

Therefore, the value of x within the interval \[{{0}^{\circ }}\le x\le {{360}^{\circ }}\] is x = \[{{120}^{\circ }},{{240}^{\circ }}\].

Note: Students make mistakes while finding the value of x within the interval. We should know some trigonometric degree values to solve these types of problems. We should also substitute the correct degree values to get the exact answer for the given problem.