Solve this trigonometric equation. 1 2 ( sin ⁡ ( θ ) + cos ⁡ ( θ ) ) = 1 2 Divide both sides by 1 2 sin ⁡ ( θ ) + cos ⁡ ( θ ) = 1Divide both sides by cos ⁡ ( θ )tan(θ)+1=sec(θ)square both sides: tan ⁡ ( θ ) + 1 = sec ⁡ ( θ ) tan 2 ⁡ ( θ ) + 2 tan ⁡ ( θ ) + 1 = sec 2 ⁡ ( θ )Use the identity sec 2 ⁡ ( θ ) = tan 2 ⁡ ( θ ) + 1: tan 2 ⁡ ( θ ) + 2 tan ⁡ ( θ ) + 1 = tan 2 ⁡ ( θ ) + 1 ∴ 2 tan ⁡ ( θ ) = 0∴ tan ⁡ ( θ ) = 0∴ θ = 0 , π , 2 πI know that 0 and 2 π are correct but that π is wrong. I also know that the other correct answer is π 2 .

Question

Solve this trigonometric equation.       1          2        (  sin  ⁡  (  θ  )  +  cos  ⁡  (  θ  )  )  =      1          2      Divide both sides by       1          2        sin  ⁡  (  θ  )  +  cos  ⁡  (  θ  )  =  1Divide both sides by   cos  ⁡  (  θ  )tan(θ)+1=sec(θ)square both sides:  tan  ⁡  (  θ  )  +  1  =  sec  ⁡  (  θ  )      tan    2    ⁡  (  θ  )  +  2  tan  ⁡  (  θ  )  +  1  =      sec    2    ⁡  (  θ  )Use the identity       sec    2    ⁡  (  θ  )  =      tan    2    ⁡  (  θ  )  +  1:      tan    2    ⁡  (  θ  )  +  2  tan  ⁡  (  θ  )  +  1  =      tan    2    ⁡  (  θ  )  +  1  ∴  2  tan  ⁡  (  θ  )  =  0∴  tan  ⁡  (  θ  )  =  0∴  θ  =  0  ,  π  ,  2  πI know that 0 and   2  π are correct but that   π is wrong. I also know that the other correct answer is       π    2  .

Freddy Doyle · Accepted Answer

The reason you got the extraneous solution   θ  =  π is because you squared both sides of the equation   tan  ⁡  θ  +  1  =  sec  ⁡  θ. You can check this by noting that   tan  ⁡  π  +  1  =  1 while   sec  ⁡  π  =  −  1, so   θ  =  π is a solution to the squared equation but not the original. On the other hand, you missed out the solution   θ  =  π      /    2 because you divided by   cos  ⁡  θ throughout, in which you&#039;ve implicitly assumed   cos  ⁡  θ  ≠  0 and hence   θ  ≠  π      /    2But these are rather easy to fix: check for extraneous solutions by substituting everything back into the original equation, and discuss the case   cos  ⁡  θ  =  0 (i.e.   θ  =  π      /    2) separately. Other than these two issues, your solution is perfect (and quite smart, actually).

Solve this trigonometric equation. 1 <msqrt> 2 </msqrt> </mfrac> ( sin

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