Trigonometry! #138

Geometry Level 3

The number of values of x ∈ [ 0 , 2 π ] x\in[0,2\pi] such that 3 sin ⁡ x + 2 ⋅ cos ⁡ x = 0 \sqrt{3\sin x}+\sqrt{2}\cdot\cos x=0 is?

This problem is part of the set Trigonometry .

0 2 4 1

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Omkar Kulkarni
Feb 21, 2015

3 sin ⁡ x = − 2 cos ⁡ x 3 sin ⁡ x = 2 cos ⁡ 2 x 3 sin ⁡ x = 2 − 2 sin ⁡ 2 x 2 sin ⁡ 2 x + 4 sin ⁡ x − sin ⁡ x − 2 = 0 2 sin ⁡ x ( sin ⁡ x + 2 ) − ( sin ⁡ x + 2 ) = 0 ( 2 sin ⁡ x − 1 ) ( sin ⁡ x + 2 ) = 0 2 sin ⁡ x − 1 = 0 sin ⁡ x = 1 2 x = 3 0 ∘ , 15 0 ∘ \sqrt{3\sin x}=-\sqrt{2}\cos x \\ 3\sin x = 2\cos^{2}x \\ 3\sin x=2-2\sin^{2}x \\ 2\sin^{2}x+4\sin x-\sin x-2=0 \\ 2\sin x(\sin x+2)-(\sin x+2)=0 \\ (2\sin x-1)(\sin x+2)=0 \\ 2\sin x-1=0 \\ \sin x=\frac{1}{2} \\ x=30^{\circ},150^{\circ}

But x = 3 0 ∘ x=30^{\circ} does not satisfy the equation. Hence we have only one solution, x = 15 0 ∘ \boxed{x=150^{\circ}} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

the same way !

Mayank Holmes - 6 years, 3 months ago

Nice solution. Just thought I'd mention that there is a typo in the 4th line, where an equals sign should instead be a minus sign.

Brian Charlesworth - 6 years, 3 months ago

the problem mentions the number of values of x, so I thought I have to find x-es which satisfy the last equation (sin x =1/2). if I read the range correctly, it says x ranges from 0 to 2 pi. so it means that x has two positive values, which are from 1st and 2nd quadrant, which are exactly x1 = 30 degree and x2 = 150 degree. I think the answer should be 2 am I missing something on this? correct me if I'm wrong

Diang Kameluh - 6 years, 3 months ago

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Good point. I'll edit my solution. In the second step, I squared both sides. This produces the chance of obtaining extra roots. It so happens that 3 0 ∘ 30^{\circ} does not satisfy the equation. Hence, 15 0 ∘ 150^{\circ} is the only solution.

Omkar Kulkarni - 6 years, 3 months ago

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