Can we solve the equation 3 cos x + 1 = 2 sin x 3\cos x + 1 = 2\sin x 3 cos x + 1 = 2 sin x
Question
Find the values of x x x 0 ∘ 0^\circ 0 ∘ 360 ∘ 360^\circ 36 0 ∘
3 cos x + 1 = 2 sin x . 3\cos x + 1 = 2\sin x. 3 cos x + 1 = 2 sin x . Suggestion
Could we rewrite sin ( x − α ) \sin (x-\alpha) sin ( x − α )
How many solutions would you expect to find in the range?
Can you sketch a graph to illustrate the equation?
Solution
We can rearrange our equation to 2 sin x − 3 cos x = 1. 2\sin x- 3\cos x = 1. 2 sin x − 3 cos x = 1.
Now we can use the
compound angle identity
to rewrite the left hand side as
2 sin x − 3 cos x = R sin ( x − α ) = R sin x cos α − R cos x sin α 2\sin x- 3\cos x = R\sin(x-\alpha) = R\sin x\cos\alpha - R\cos x\sin\alpha 2 sin x − 3 cos x = R sin ( x − α ) = R sin x cos α − R cos x sin α and matching the coefficients tells us we need
R R R
and
α \alpha α
such that
R cos α = 2 and R sin α = 3. R\cos\alpha=2 \quad\text{and}\quad R\sin\alpha=3. R cos α = 2 and R sin α = 3. This reduces to
R 2 = 13 and tan α = 3 2 ⟹ R = 13 and α = 56.3 ∘ ( 1 d.p. ) \begin{align*} R^2=13 \quad&\text{and}\quad \tan\alpha=\frac{3}{2} \\ \implies\quad R=\sqrt{13} \quad&\text{and}\quad \alpha=56.3^\circ (1 \text{d.p.}) \end{align*} R 2 = 13 ⟹ R = 13 and tan α = 2 3 and α = 56. 3 ∘ ( 1 d.p. ) So now our original equation becomes sin ( x − 56.3 ∘ ) = 1 13 . \sin(x-56.3^\circ)=\frac{1}{\sqrt{13}}. sin ( x − 56. 3 ∘ ) = 13 1 .
The general solution to this is x = arcsin 1 13 + 56.3 ∘ or x = 180 ∘ − arcsin 1 13 + 56.3 ∘ x=\arcsin\frac{1}{\sqrt{13}}+56.3^\circ \quad\text{or}\quad x=180^\circ - \arcsin\frac{1}{\sqrt{13}}+56.3^\circ x = arcsin 13 1 + 56. 3 ∘ or x = 18 0 ∘ − arcsin 13 1 + 56. 3 ∘ 360 ∘ 360^\circ 36 0 ∘
Within the required range this gives us x = 72.4 ∘ or 220.2 ∘ ( 1 d.p. ) . x=72.4^\circ\quad\text{or}\quad 220.2^\circ (1 \text{d.p.}). x = 72. 4 ∘ or 220. 2 ∘ ( 1 d.p. ) .
Check – if we have access to a graphing program, we can see where the graphs of y = 3 cos x + 1 y = 3\cos x + 1 y = 3 cos x + 1 y = 2 sin x y = 2\sin x y = 2 sin x
