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

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

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Find 𝜃 in degrees given sec (180° + 𝜃) = −(2√(3))/3 where 𝜃 is the smallest positive angle.

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Video Transcript

Find 𝜃 in degrees given sec of 180 degrees plus 𝜃 is equal to negative two root three over three where 𝜃 is the smallest positive angle.

In order to solve this problem, we recall that sec 𝜃 is equal to one over cos 𝜃. It is the reciprocal of cos 𝜃. We can therefore rewrite the equation as one over cos of 180 plus 𝜃 is equal to negative two root three divided by three. If two fractions 𝑎 over 𝑏 and 𝑐 over 𝑑 are equal, then their reciprocals must also be equal. This means that in our question, cos of 180 degrees plus 𝜃 is equal to negative three over two root three. Rationalizing the denominator of the right-hand side by multiplying the top and bottom by root three gives us negative root three over two. This is equal to cos of 180 degrees plus 𝜃.

We recall that cos of one of our special angles, 30 degrees, is equal to root three over two. Using our CAST diagram, we can find the angles where cos of 𝜃 is equal to negative root three over two. These will be equal to 180 minus 𝜃 and 180 plus 𝜃. cos of 150 degrees and cos of 210 degrees are equal to negative root three over two. This means that 180 plus 𝜃 is equal to 150, and 180 plus 𝜃 is equal to 210. Subtracting 180 from both sides of the equations gives us 𝜃 equals negative 30 and 𝜃 equals 30.

We are told in the question that 𝜃 must be a positive angle. The smallest positive solution of sec of 180 degrees plus 𝜃 is equal to negative two root three over three is 30 degrees.

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