Question Video: Solving Equations Using Inverse Trigonometric Functions | Nagwa Question Video: Solving Equations Using Inverse Trigonometric Functions | Nagwa

Question Video: Solving Equations Using Inverse Trigonometric Functions Mathematics

Given that π‘₯ is an acute angle and 4 cos (π‘₯) = 2√3, determine the value of π‘₯ in radians.

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Video Transcript

Given that π‘₯ is an acute angle and four multiplied by the cos of π‘₯ equals two root three, determine the value of π‘₯ in radians.

We are told that π‘₯ is an acute angle, which means it lies between zero and 90 degrees. However, in this question, we want our answer in radians. We recall that 180 degrees is equal to πœ‹ radians. This means that 90 degrees is equal to πœ‹ over two radians and π‘₯, therefore, lies between zero and πœ‹ by two. We will begin to solve the equation in this question by making cos of π‘₯ the subject. We can do this by dividing both sides of the equation by four. The left-hand side simplifies to the cos of π‘₯. Dividing the numerator and denominator of the right-hand side by two gives us root three over two. The cos of π‘₯ equals root three over two.

We can then solve this equation using our knowledge of the inverse trigonometric functions. We know that for any acute angle πœƒ, the inverse cos of cos πœƒ is equal to πœƒ. Taking the inverse cosine of both sides of our equation gives us the inverse cos of cos π‘₯ is equal to the inverse cos of root three over two. Since π‘₯ is an acute angle, we can therefore conclude that π‘₯ is equal to the inverse cos of root three over two.

We can then type the right-hand side into our calculator. Ensuring that we’re in radian mode, we get an answer for π‘₯ of πœ‹ over six. If π‘₯ is an acute angle and four multiplied by the cos of π‘₯ is two root three, then π‘₯ is equal to πœ‹ over six radians. It is worth noticing here that we could’ve solved the final step of our equation using our knowledge of special angles. We recall that the cos of 30 degrees is equal to root three over two. This means that the cos of πœ‹ over six radians is also equal to root three over two. Taking the inverse cosine of both sides of this equation, we see that πœ‹ over six is equal to the inverse cos of root three over two. This confirms that we have the correct answer for π‘₯.

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