Question Video: Solving Trigonometric Equations Involving Special Angles and Reciprocal Identities | Nagwa Question Video: Solving Trigonometric Equations Involving Special Angles and Reciprocal Identities | Nagwa

# Question Video: Solving Trigonometric Equations Involving Special Angles and Reciprocal Identities Mathematics • First Year of Secondary School

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Find the set of values satisfying 2 sin 𝜃 + cos 𝜃 sec 𝜃 = 0 where 0° ≤ 𝜃 < 360°.

02:14

### Video Transcript

Find the set of values satisfying two sin 𝜃 plus cos 𝜃 sec 𝜃 equals zero, where 𝜃 is greater than or equal to zero and less than 360 degrees.

In order to solve this equation, we begin by recalling that sec 𝜃 is equal to one over cos 𝜃. It is the reciprocal of cos 𝜃. We can, therefore, rewrite the equation as two sin 𝜃 plus cos 𝜃 multiplied by one over cos 𝜃 is equal to zero. The cos 𝜃’s cancel such that two sin 𝜃 plus one is equal to zero. We can then subtract one from both sides of this equation. Finally, dividing by two gives us sin 𝜃 is equal to negative one-half. We recall that sin of 30 degrees is one of our special angles and is equal to one-half.

By drawing our CAST diagram, we can see that the solutions for sin 𝜃 equals negative one-half will be between 180 and 270 degrees and also between 270 and 360 degrees. The angles these lines make with the horizontal will be equal to 30 degrees. One of our solutions will be equal to 180 plus 30. This is equal to 210. The second solution will be equal to 360 minus 30. This is equal to 330. The values of 𝜃 that satisfy the equation are 210 degrees and 330 degrees.

This can be written using set notation as shown. The set of values that satisfy two sin 𝜃 plus cos 𝜃 sec 𝜃 is equal to zero are 210 degrees and 330 degrees.

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