Solving cos 2 ⁡ θ − sin ⁡ θ = 1 , θ ∈ [ 0 ∘ , 360 ∘ ]

Question

Solving       cos          2        ⁡      θ    −  sin  ⁡      θ    =  1  ,    θ  ∈  [      0          ∘        ,      360          ∘        ]

iskakanjulc · Accepted Answer

Notice, we have      cos    2    ⁡  θ  −  sin  ⁡  θ  =  1  1  −      sin    2    ⁡  θ  −  sin  ⁡  θ  =  1  −      sin    2    ⁡  θ  −  sin  ⁡  θ  =  0      sin    2    ⁡  θ  +  sin  ⁡  θ  =  0  sin  ⁡  θ  (  sin  ⁡  θ  +  1  )  =  0  sin  ⁡  θ  =  0    ⟺    θ  =  n  (      180    ∘    )Where, n is any integer.But for the given interval   [      0    ∘    ,      360    ∘    ], substituting n=0,1,2, we get  θ  =      0    ∘    ,      180    ∘    ,      360    ∘  Now,  sin  ⁡  θ  +  1  =  0  sin  ⁡  θ  =  −  1    ⟺    θ  =  2  n  (      180    ∘    )  −      90    ∘  But for given interval   [      0    ∘    ,      360    ∘    ], substituting n=1 we get  θ  =      270    ∘  Hence, we have      θ    =      {                  0        ∘            ,              180        ∘            ,              270        ∘            ,              360        ∘              }

2d3vljtq · Accepted Answer

cos          2        ⁡  (  θ  )  −  sin  ⁡  (  θ  )  =      cos          2        ⁡  (  θ  )  +      sin          2        ⁡  (  θ  ) , (using Pythagorean Identity)      sin          2        ⁡  (  θ  )  +  sin  ⁡  (  θ  )  =  0let   x  =  sin  ⁡  (  θ  )      x          2        +  x  =  0  x  =            −      1      ±                        1                      2                          −        4        (        1        )        (        0        )              2  , (Quadratic Formula)x=0, x=−1substitute   sin  ⁡  (  θ  ) back in and solve  0  =  sin  ⁡  (  θ  )  ⇒  θ  =      0          ∘        ,      180          ∘        ,      360          ∘        −  1  =  sin  ⁡  (  θ  )  ⇒  θ  =  −      270          ∘        θ  =  {      0          ∘        ,      180          ∘        ,      270          ∘        ,      360          ∘        }

Solving cos <mrow class="MJX-TeXAtom-ORD"> 2 </mrow> </msup> &#x206

Answered question

Answer & Explanation

New Questions in Precalculus